Heat Transfer

The term represents the partial derivative of Q (or the amount of heat transfer) with respect to time, while represents the absolute derivative of Q with respect to time. The the amount of heat transfer (Q) depends not only on time but also on the factors such as the area of cross-section, temperature difference, etc. So, we cannot write it as a complete derivative of time only. Hence the better way of representation of heat current is as a partial derivative or in the given form, which is, .

Yes, the body will radiate at 20°C. Every entity with a temperature greater than 0K will radiate. In this case, as the room temperature(30°C) is greater than the body temperature, the temperature of the body would not fall further. Since the surrounding is warmer, the rate of absorption of radiation of the body from the surrounding will be comparatively greater than the thermal loss of the body by its own radiation. As a result, the temperature of the body will does not further fall.

The cooling down of tea is facilitated by Convection more than evaporation or radiation. Since the tea is a hot body, it does radiate and evaporate. But when we blow over the hot tea, convection plays the major role in cooling the tea down.

When we blow, the cool air of wind will replace the hot air from the upper part of the hot tea. This cool air will go down near to the tea because of the higher density compared to hot air. This will eventually make the tea cool down faster.

No, This will rather cause an increase in room temperature than decreasing it.

The refrigerator is a device which takes the heat from the refrigerator cabin and expels it outside the fridge, with the help of external work. If we keep the refrigerator door opened while keeping the windows and doors closed, the refrigerator will take in the heat from the room and expel it into the room itself. Since no instrument can work with 100% efficiency, the loss in the electrical energy of the fridge will dissipate as heat, and this along with the heat after refrigeration, will be expelled into the room itself. Hence, in a room with an opened fridge door, it will expel more heat than it takes in from the room.

Since all the windows of the room are closed, opening the refrigerator door will result in an increase in the room temperature instead of decreasing it.

It would be better to choose the wooden chair over the metal chair, on a winter night. This is because of the fact that wooden chair transfers less heat into the human body than the metal chair. These behaviors can be quantified by the difference in the conductivity between the two materials of the chairs. Metal has more thermal conductivity than wood. Hence metal chair will tend to transfer more heat or cold to one’s body while in contact with it.

This can also be explained by the concept of specific heat capacity. The specific heat capacity of metal is lower than that of wood. Hence, it is easier to increase the temperature of the metal, by one degree, compared to that of wood. So, when the human body comes in contact with a metal and a wood, both having the same lower temperature than the human skin, the metal will absorb heat at a faster rate than wood. This will result in the cooling of the human body. Since the metal chair with a lower temperature has a faster rate of heat transfer, the wooden chair will be comfortable on a winter day.

So, on a winter day, it is preferable to sit in the wooden chair, which transfers comparatively less cold, hence helps to avoid the high cold of the surroundings.

Yes, both the metal balls will radiate since their temperature is greater than 0 K. The radiation heat transfer is irrespective the surrounding medium.

The thermal radiation from a body depends on its temperature, according to Stefan–Boltzmann law. The law states that the radiation heat transfer from a black body is proportional to the fourth power of its absolute temperature ().

Or,

Where,

Radiation heat transfer

Stefan–Boltzmann constant 5.67 x 10^-8 W/m² k⁴

Area of the black body

Hence, the rate of radiation heat loss from the body at 300K will be less compared to the rate of radiation heat loss from the body at 600K. Since the bodies are in vicinity of each other and the radiation is happening in all the directions, the interactions of thermal radiation can cause the equalization of the temperatures on the balls.

Also, the heat loss from the hotter ball to the cooler ball or the heat gained by the cooler ball will be proportional to according to Stefan’s law.

An electric fan does not cool the air. It just creates a forced circulation of the air present inside the room. This circulation will facilitate the evaporation of sweat, that is present on the surface of one’s skin. The heat for the evaporation is taken from the skin itself, and this will cause a sense of cooling.

Hence, on a summer day, even though the air is not getting cooled by the fan, the evaporation of the sweat will create comfort for the body.

At high altitude, even though the temperature is higher, the density of the air surrounding an animal is very low. This will cause a drastic decrease in the conduction mode of heat transfer, as the heat transfer occurs due to the close interaction of the molecules. So, there is very less amount of molecule in the atmosphere to transfer the heat to the animal’s body. But, the animal skin contains water on the surface and the boiling point of this water will get decreased drastically due to the extreme drop of pressure in the higher altitude; and thus, water will take the heat from inside the body, instead of taking it from the atmosphere. This phenomenon will cause to drop the temperature inside the animal body.

Because of this lack of heat transfer from the surrounding air, the animal body will get freezed instead of getting heated.

Yes. The temperature in an open place will be higher than that of a shady place due to the presence of more number of heated molecule in the former case.

Human body needs thermal comfort despite the surrounding condition. On a cold environment, human body needs to transfer less amount of heat compared to that in a normal environment.

The temperature of the surrounding is lower than the usual temperature on a winter day, and human body will transfer more heat to surrounding by radiation, convection and conduction*, causing a drop in the temperature of the outer skin. In an open space, where there is more radiation from sun, the temperature of the surrounding will be comparatively higher than a shady place. So, the heat transfer from human body will be lower in the open space since all the above modes of heat transfers are directly proportional to the Temperature difference between human body and the surroundings.

In the case of shady place, the temperature of the surroundings will be comparatively lower due to 2 reasons;

1. There is lesser availability of direct radiation from the sun

2. The phenomenon of transpiration** from the leaves of the trees will give a cooling effect.

So, there will be a higher amount of heat transfer from human skin to the colder surroundings.

To conclude, on a winter day, a hot environment will be convenient in order to reduce the heat transfer from the body.

*There are other types of heat transfer such as the heat transfer caused by Breathing, etc.

**Transpiration is the process of removal of water from leaves through the pores called Stomata. The evaporation of this will cause a cooling effect.

The temperature of the environment is warm when the air gas molecules get warm. The accumulated heat from the sun, mainly by radiation, on the day times are emitted back from the earth at nights.

The earth’s surface radiate energy in the form of Infrared radiation at night. This will attribute to the heating of air molecules in the environment. However, the heat from the radiation is usually gone into space; But if the clouds are present in the sky, they will trap the infrared radiations and prevent the heat from leaving the environment. As clouds are a form of water vapors, they have a tendency to absorb more heat compared to air molecules. This heat will be emitted back to the earth surface from the clouds, which causes the temperature to rise further. While in the case of clear sky, there is nothing to prevent radiation from escaping from the environment.

The comfort of wearing a white dress over a dark colored dress on a summer day is due to the difference in absorption of the sun’s light by different colors.

The light is an electromagnetic wave, that contains mass and energy. Based on the absorption of certain wavelengths of light energy source like Sun, objects will exhibit its color. A darker object is the one which absorbs all the colors (the wavelengths in the visible spectrum) of the light and a white object will reflect all the colors of the light.

Hence, a dark-colored dress will absorb all the visible wavelengths from the sun and this energy will get converted into heat energy. While in the case of a white dress, all the visible wavelengths are reflected from the dress causing effectively no absorption (in the ideal cases).

So, in the summer, as the radiation from the sun is higher, wearing a white dress will help to not absorb the energy but to reflect it, while a darker dress will absorb most of the radiation. Hence it would be more comfortable in a white dress on a summer day.

The Thermal conductivity of rod depends on the material of the rod. Heat transfer is due to free electrons and conductors like metals have electrons in their outer shell so as to move freely along the road. If the rod is made up of non-conducting material then the transfer of heat would not pe possible as they don’t have free electrons .
Thermal conductivity does not depend on mass as the mass of the rod remains the same and it doesn’t affect the number of free electrons in the rod.
Thermal conductivity does not depend on length and area of cross section as thermal conductivity is the ability of a material to conduct heat and the dimensions of the material does not affect its ability.
Thus, option (d) is the correct option.

Transfer of heat due to conduction is the molecular vibrations which does not involve mass movement. In the room, heat conduction is possible due the molecular collision of air molecules and other molecules present.
Transfer of heat due to convection is due to actual movement of heated material or molecules. In the room, the movement of air molecules from one place to another can result in heat transfer.
Transfer of heat due to Radiation does not need any medium or material. So, in a room containing air energy is radiated by every body or object, thus heat transfer is possible.
Thus, option (d) is the correct option.

Stefan- Boltzmann Law is used.
According to the law, the energy of thermal radiation emitted per unit time by a body having surface area A is given as:

Here, e is the emissivity of the body and σ is the Stefan-Boltzmann constant and T is the Temperature.
As , temperature of solid will increase.
Now, for the solid the energy of thermal radiation is:

Energy of the thermal radiation in evacuated chamber is:

Now, the net difference in energy is .
Hence the net energy difference is proportional to:

Hence, The rate of increase of temperature of the body is proportional to .
For options (a), (b) and (c) the power od temperature difference is not 4. Hence, they are incorrect.
Thus, option (d) is the correct option.

The thermal radiation emitted by a body in proportional of Tⁿ where T is its absolute temperature. The value of n is exactly 4 for all bodies.
According to Stefan-Boltzmann Law:
The energy of thermal radiation emitted per unit time by a body having surface area A is given as:

Here, e (between 0 to 1) is the emissivity of the body and σ is the Stefan-Boltzmann constant and T is the Temperature.
For black body, e=1.
Thus, option (b) is the correct option.

The thermal radiation emitted in a given time by A and B are in the ratio 1: 1.15.
Stefan-Boltzmann Law is given as:
The energy of thermal radiation emitted per unit time by a body having surface area A is given as:

Here, e (between 0 to 1) is the emissivity of the body and σ is the Stefan-Boltzmann constant and T is the Temperature.
Now the temperature should be in Kelvin.
For body A :
For body B :
Here T_A = 273 + 10 ° C = 283 K.
T_B = 273 + 20° C= 293 K
Substituting we get,


Out of all the options only 1:1.15 gives answer close to 0.8703 hence,

Thus option (a) is the correct option.

In steady state, the temperature at any point of the material remains unchanged as time passes. Hence the end near the furnace will be extremely hot whereas the other end will have minimum temperature. Thus, showcasing nonuniformity in the temperature throughout the metal rod.
In steady state there is no change in temperature hence first two options are not valid.
The temperature us not constant throughout the rod, it’s different at different points. Hence third option is incorrect too.
Thus option (d) is the correct option.

Newton’s law of cooling is a special case of Stefan’s law.
Stefan’s Law states that:
The energy of thermal radiation emitted per unit time by a body having surface area A is given as:

Now, the energy absorbed by a body due to radiations emitted by the walls per unit time is given as:

Here T₀ is the temperature of the surrounding.
Now the net rate of loss of energy of the body due to thermal radiation is:


If the temperature difference is small, we can write



Using Binomial Expansion for the forth power we get and using equation relating heat and specific heat we get,

For a small difference in temperature of the body and the surroundings, the rate of cooling is directly proportional to the Area exposed of the body and the temperature difference.
It can also be written as,

This is Newton’s Law of Cooling. b is a constant and depends on the nature of the surface and surrounding conditions.
Negative sign shows that the temperature decrease with time.
Hence. We started off with Stefan’s Law and Ended with Newton’s Law of cooling.
Thus, option (c) is the correct option.

The temperature of the hot liquid will decrease with time. The walls of the room will absorb the emitted radiation from the liquid. Since the room is big the temperature difference between emitted radiation and absorbed radiation is large.
According to Stefan’s Law, the energy of thermal radiation emitted per unit time is directly proportional to T⁴. Thus, decrease in temperature with time will also be proportional to T⁴. Thus the curve must be exponentially decreasing curve which is curve a.
Thus, curve a represents the plot.

When a hot liquid is kept in a big room, the temperature of the hot liquid will decrease with time. The room is big hence the difference between emitted radiation and absorbed radiation by the walls would be large. According to Stefan’s Law, the energy of thermal radiation emitted per unit time is directly proportional to T⁴. The Temperature would show an exponentially decreasing behavior against time. After a sometime, temperature of the liquid becomes equal to the temperature of the surrounding resulting in zero temperature difference. When logarithm of temperature difference is taken, the fourth power becomes coefficient as per the logarithmic properties,

Here T is the emitted radiation and T₀ is the absorbed radiation. Thus, the logarithmic plot would be a straight line as logarithmic graphs are nonlinear variables converted to linear graphs.
Thus, option (a) is the correct option.

According to Newton’s Law of Cooling:

Here is the rate of fall of temperature is the constant depending on nature of surface and surrounding conditions, A is the area of cross-section of the body, T is the average Temperature of the body and T₀ is the surrounding temperature.
It can also be written in Celsius scale as

Where θ and θ₀ are the temperature of the body and surrounding in Celsius scale respectively , k is a constant.
First condition:
Temperature falls from 65° C to 60° C.
Which means 5°C falls in 5 minutes,
∴
θ = 62.5 ° C
Substituting we get,


Second Condition:
Temperature further falls from 60 °C to 55 °C.
∴ dθ= 5 °C.
θ =
Substituting value of -k,dθ and θ in the equation we get,


Here we can conclude that time required for the body to go from 60 ° C to 55° C will be,


Thus, option (c) is the correct option.

When one end of the metal rod is dipped in boiling water, the temperature at that end would be maximum. When the other end of the rod is dipped in melting ice, the temperature of that end would be minimum. Due to conduction, the temperature of the rod starts to increase from maximum temperature end to minimum temperature end. After some time, equilibrium is established with the surrounding.
When Steady state is reached, the temperature throughout the rod is nonuniform and constant. Temperature will be different at different points on the rod. Due to equilibrium the state of the rod does not change.
Thus, option (d) is the correct option.

A blackbody is a body which absorbs all incident radiation falling on it and the radiation emitted by it is called as Blackbody Radiation.
A 100% Blackbody doesn’t Reflect as it absorbs all the incident radiation neither does it refracts radiation as it does not have refractive properties.
Lampblack is close to a black body, it reflects about 1% of the incident radiation
Thus, options (c) and (d) are correct options.

In summer, the shore is warmer than the river. Hence air above the shore is at high temperature compared to the temperature of the air above of the river. The warmer air particles flow from river to the shore due to convection. Therefore, there is a mild wind at the shore of a calm river. Whereas during winters it’s opposite. The temperature of the air above the river is warmer compared to that of air above land. Thus, convection current flows from land to river.
Thus, option (b) is the correct option.

Steel is a good conductor of heat whereas charcoal isn’t. So, the sunlight falling on the steel surface will make the surface hot and since charcoal is an absorber the incident sunlight would be absorbed by the charcoal. Since charcoal is bad conductor of heat, it won’t be too hot. Therefore, if both are picked up by bare hands, the steel will be felt hotter than the charcoal.
Secondly, Blackbodies are good absorbers of radiation and hence good emitters of radiation. Charcoal is a black body and thus it is a good emitter compared to steel. Hence, If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel.
Option (a) is incorrect as steel is a bad absorber of heat compared to charcoal as charcoal is a blackbody.
Option (b) is incorrect as temperature is the intensity of heat present in the substance. Steel is hot at the surface but charcoal has more heat intensity absorbed within it.
Thus, options (c) and (d) are the correct options.

According to Wein’s Displacement Law:
The wavelength of the peak of the blackbody radiation is inversely proportion to the absolute temperature of the emitter.

Here, λ_m is the wavelength at maximum intensity, T is the absolute temperature in kelvins and b is the Wein constant.
Also,

Where c is speed of light and v is the frequency.
Now when the absolute temperature of the body is doubled,
T becomes 2T. Since LHS is equal to a constant, the RHS of the Wein’s displacement law must be doubled too.
This becomes,

Here v₀ is the frequency at maximum intensity. As we can see doubling absolute temperature doubles frequency too.
Hence option (a) is correct.
Now, According to Stefan’s Law, the rate of emission of radiated energy is proportional to the fourth power of absolute Temperature,


where u is the radiated energy. Thus, the total energy emitted will increase by a factor of 16.
Thus, options (a) and (c) are the correct options.

According to Stefan- Boltzmann Law:
The energy of thermal radiation emitted per unit time by a body having surface area A is given as:

Here, e (between 0 to 1) is the emissivity of the body and σ is the Stefan-Boltzmann constant and T is the Temperature.
Since the spheres have equal radii and are of same material, the area of cross section would be same for both the spheres.
Thus Stefan’s Law will hold true for both the spheres in the exact same way. Hence, both will emit equal amount of radiation per unit time in the beginning.
The energy of radiation absorbed by the spheres should be equal to the energy of radiation emitted by them. Applying same Stefan-Boltzmann
Law we get,

Where u₀ is the energy of radiation absorbed and T₀ is the initial surrounding temperature.
Again due to equal Area of cross section and same material, both will absorb equal amount of radiation from the surrounding in the beginning.
Thus, options (a) and (b) are the correct options.

Given:
Area of the uniform slab: A = 10× 10 cm² = 100 cm² = 0.01 m²
Height of the slab : x = 1 cm = 0.01 m
Temperature difference of two heat reservoirs:
ΔT = 90-10=80° C
The thermal conductivity of the material: K=0.80 W m^–1 °C^–1.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred , ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Substituting we get,




Hence, the amount of heat flowing through slab is 3840 J/min.

Given:
Thickness of the container : x = 1 cm = 0.01 m
Thermal conductivity of the sheet : K= 0.025 J s^–1m^–1°C^–1
Temperature of the liquid nitrogen: T₁ = 80 K
Area of the container : A = 0.8 m²
Temperature of the atmosphere : T₂ = 300 K
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Here, ΔT = T₂- T₁ = 220 K
Substituting the values we get,


Hence, the rate of heat flow from the atmosphere to the liquid nitrogen is 440 J/s.

Given:
Body Temperature : T₁ = 97 °F = 36.1 °C
Room temperature: T₂ = 47°F = 8.3 °C
Surface area under clothes : A = 1.6 m²
Thermal conductivity of the cloth : K =0.04 J s–¹ m^–1 °C^–1
Thickness of the cloth : x = 0.5 cm = 0.005 m
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Here, ΔT = T₁-T₂ = 36.1-8.3 = 27.8 °C
Substituting we get,


Hence, the rate at which heat is flowing out of his body through the clothes is 355.84 J/s.

Given:
Area of the bottom of container: A = 25 cm² = 0.0025 m².
Thickness of the container: x = 1 mm = 0.001 m
Thermal conductivity of the container: K =50 W m^–1 °C^–1.
Latent heat of vaporization : L = 2.26 × 10⁶ J kg^–1.
Mass of the water converted to steam : m = 100 g = 0.1 kg
Temperature of the water : T₂ = 100 ° C
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
Also, Δt = 1 minute = 60 seconds
Let temperature of the bottom of the container be T₁.
Substituting we get,



∴
∴
∴
Hence, Temperature of the bottom of the container is
130.13 ° C.

Given:
Thermal conductivity of the steel rod : K=46 J s^–1 m^–1 °C^–1.
Length of the rod : x = 1 m
As heat flows from area of high temperature to low temperature,

Temperature of the end in water : T₁ = 100° C
Temperature of the end in ice : T₂ = 0° C
Area of cross section of the rod :
A = 0.04 cm² = 0.04× 10^-4 m².
Latent heat of fusion of ice = 3.36 × 10⁵ J kg^–1.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
Here, Δt = 1 second.
Substituting,



∴ m = 5.47 × 10^-8 kg
Hence, mass of the ice melting per second is 5.47 × 10^-8 kg.

Given:
Temperature of the water : T₁ = 20° C
Temperature of the ice box : T₂= 0° C
Surface area of the box : A = 2400 cm² = 0.24 m²
Thickness of the box : x = 2 mm = 0.002 m
Thermal conductivity of the box : K = 0.06 W m^–1 °C^–1.
Latent heat of fusion of ice = 3.4 × 10⁵ J kg^–1.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
We need to find rate at which ice melts, which means we need to calculate the decrease in mass of the ice per second: Δm/Δt
Where Δm is the rate of change of mass.
Now,


Substituting for Δθ we get



OR

Hence, the rate at which ice melts in the box is
4.23× 10^-4 kg/s or 1.52 kg/hr.

Given:
Thickness of the pitcher: x = 1 mm = 0.001 m
Mass of the water in the pitcher : m = 10 kg
Rate at which water evaporates at it’s outer surface:
= 0.1 gs^-1 = 0.1× 10^-3 kg s^-1.
The surface area of the pitcher : A = 200 cm²= 0.02 m².
Room temperature : T₁ = 42 ° C
Latent heat of vaporization: L = 2.27 × 10⁶ J kg^–1.
Thermal conductivity of the porous walls:
K = 0.80 J s^–1m^–1 °C^–1.
Let the constant temperature of the water in the pitcher be T_2.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
Now, 0.1× 10^-3 kg of water evaporates in 1 second. Thus by unitary method, 10 kg of water will evaporate in 10⁵ seconds.
Δt = 10⁵ seconds.
Substituting we get,



∴ 42 – T₂ = 14.18
∴ T₂ = 42 - 14.18
∴ T₂ = 27.82 °C
Hence, the temperature of water in the pitcher when it attains a constant value is 27.82 °C.

Given:
Thermal conductivity of steel frame: K=45 W m^–1 °C^–1.
Length of the steel frame : x = 60 cm = 0.6 m
Area of cross section : A = 0.20 cm² = 0.2 × 10^-4 m².
Temperature difference between the free ends :
ΔT = 40-20= 20°C.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Substituting the values we get,


Hence, the rate of heat flow through a cross-section of the frame is 0.03 J/s.

Given:
Temperature of the water : T₁ = 50 °C
Height of the vessel : h = 10 cm = 0.1 m
Cross section area of the vessel: A = 10 cm² = 0.001 m².
Thickness of the flat parts: x = 1mm = 0.001 m
Thermal conductivity of the aluminium: K =200 J s^–1 m^–1 °C^–1
Temperature outside: T₂ = 20 °C
Density of water : ρ = 1000 kg m^-3
The specific heat capacity of water = 4200 J k^–1 m^–1 °C^–1.

Since the walls are adiabatic, no heat transfer would take place. Hence the heat would be transferred only via flat surfaces which are Up and Bottom surfaces.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Rate of flow of heat from both the flat surfaces would be:



Hence, rate of heat flow from both the flat surfaces is 12000 J/s.
We know that,
ΔQ=Δθ = msΔT

Where, ΔQ is the change in heat energy, m is the mass , s is the specific heat of the substance and ΔT is the change in temperature for the substance.
Also,
Mass = Density × Volume
∴ m = 1000 × 0.001× 0.1
∴ m = 0.1 kg
We need to find time taken for the temperature to fall by 1°C
Thus, ΔT = 1°C
Substituting in equation (2) and using (1) we get,



Hence, it took around 0.035 seconds for the temperature to drop by 1°C.

Given:
Length of the rod: x = 20 cm =0.2 m
Area of cross section of the rod: A = 0.2 cm² = 0.2× 10^-4 m².
Temperature at left end : T₂ = 20° C
Temperature at right end : T₁ = 80° C
Thermal conductivity of copper: K = 385 W m^–1 °C^–1.

Distance of point C from left end : x’ = 11cm = 0.11 m
Formula used:
(b)
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Transfer of heat due to entire rod is,


Hence, heat current in the rod is 2.31 J/s
(a)
Let T be the temperature at point C.
T>T₂ as heat flows from High temperature to low temperature.
Substituting we get,





Hence, temperature at a distance of 11 cm from the left end is 53 °C.

Given:
Temperature difference between the ends of the meter stick AB:
ΔT = T₂-T₁ = 100-0 = 100 °C
Temperature of one end of the rod: T₃ = 25 °C
Length of the rod : l = 1 m

Here, C is the point at which the other end of the rod is placed.
Distance between A and C = x
Distance between C and B = 1-x

Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Now, for zero heat current in the rod, the temperature difference must be zero: ΔT = 0.
Since one end of the rod is maintained at 25 °C, the other end must be maintained at 25 °C.
Hence heat current between A and C must be equal to the heat current between C and B


Here (ΔT)_AC and (ΔT)_CB is the temperature difference between AC and BC respectively.

∴ 75(1-x) = 25x
∴ 75 -75x = 25x
∴ 75 = 100x
∴ x = 75/100
∴ x = 0.75 m
Hence, in order to have zero heat current through the rod the other end of the rod must be placed at a distance of 0.75 m from the end at 100 ° C.

Given:
Volume of the box : V = 216 cm³ = 216× 10^-6 m³.
Thickness of the box: x = 0.1 cm = 0.001 m
Power of the heater : P = 100 W
Temperature difference : ΔT = 5 °C

Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Volume of the cube is a³. Where a is the side of the cube.
∴ a = (216× 10^-6)^1/3 = 0.06 m
As heat will be transferred from all the sides of the cube,
Surface area of the cube is : A = 6a²
∴ A = 6× (0.06)² = 0.0216 m².
We know that,
Power = Energy per unit time
Thus,

Substituting we get,

Given:
Thickness of the container: x = 2 mm = 0.002 m
Thermal conductivity of the container: K= 0.50 W m^–1 °C^–1.
Temperature of the water: T₁ = 1 °C
Temperature of the ice bath: T₂ = 0 °C
Surface are in contact with the water: 0.05 m²
Speed of the block: v = 10 cm/s = 0.1 m/s
g = 10 m/s²
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
The effect of the block going down and the heat transfer has one identity in common, which is power.
Power due to a block of mass M moving with constant velocity is given as


Here W is the work done by the block.
∴ P = (mg).v
∴ P = M× 10× 0.1
∴ P = M× 1
Also in terms of Heat Energy: P= Energy per unit time




Hence, the mass of the block is 12.5 kg.

Given:
Temperature of the water: T₁ = 0 °C
Temperature of the atmosphere: T₂ = -10 °C
Change in temperature : ΔT = T₁-T₂ = 10 °C
Length of the ice formed : l = 10 cm = 0.1 m
Density of water: ρ = 1000 kg m^–3
Latent heat of fusion of ice: L = 3.36 × 10⁵ J kg^–1.
Thermal conductivity of ice: K0. = 1.7 W m^–1 °C^–1.
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
And,
Mass = Density × Volume
∴ m = ρ V = ρAl
Here ρ is the density of water, A is the area and l is the length.
(a)
Let change in thickness be Δx,thus rate of change of thickness is:





Hence, the thickness of the ice increases at a rate of 5.05× 10^-7 m/s.
(b)
Consider time required to form thin layer of ice dx is dt.
Using above formulations,
Mass of dx: dm = Adx × ρ
And heat absorbed by the thin layer,
dQ = dm× L
Now, rate of Heat transfer due to thin layer becomes:




Integrating on both sides and setting the limit of ice formed x: 0 to 0.1






Hence, it took 27.45 hours to form 10 cm thick ice.

Given:
Temperature at the bottom of the lake: T₁ = 4 °C
Temperature above the surface : T₂ = - 10 °C
Depth of the lake: d = 1 m
Thermal conductivity of water: K_W = 0.50 Wm^–1 °C^–1.
Thermal conductivity of ice: K_I = 1.7 W m^–1 °C^–1.


Distance AB is : x_AB = x m
Distance CB is : x_CB = (1-x) m
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
In the diagram, point B depicts the maximum level upto which ice can be formed inside the lake.
Temperature at B : T₃ = 0°C
This ice attains a steady state maximum level. Steady state means that the temperature at any point remains unchanged.
This means that the temperature difference between points A,B and C would be unchanged.

This means that the rate of heat transfer between A and B equals the rate of heat transfer between B and C.






Hence, after attaining steady state the thickness of the ice below the lake is 0.894 m.

Given:
Length of all the rods: x= AB=BC=AC = 20 cm = 0.2 m
Area of cross section of these rods: A = 1 cm² = 0.0001 m².
Thermal conductivity of rod AB : K_AB = 50 J s^–1 m^–1 °C^–1
Thermal conductivity of rod BC : K_BC = 200 J s^–1 m^–1 °C^–1
Thermal conductivity of rod AC : K_AC = 400 J s^–1 m^–1 °C^–1
Temperature at A : T₁ = 40 °C
Temperature at B : T₂ = 80 °C
Temperature at C : T₃ = 80 °C
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
(1)
Rate of heat flowing in the rod AB is



Hence, rate of heat flowing through the rod Ab is 1 J/s or 1 W
(2)
Rate of heat flowing in the rod BC is



Hence, the rate of heat flowing through the rod BC is 0 as both the ends of the rods are maintained at same temperature.
(3)
Rate of heat flowing in the rod BC is



Hence, the rate of heat flowing through the rod AC is 8 J/s.

Given:

Temperature at junction 1: T₁
Temperature at junction 2: T₂
Length of the rod 1: x= d=2r
Where d is the diameter and r is the radius.
Length of the rod 2: x’ = circumference = πr
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
As both the rods have same material, their thermal conductivity is the same.
(1)
Rate of heat flowing through rod 1

(2)
Rate of heat flowing through rod 2

Ratio of the rate of heat transferred from semi circular rod to straight rod is



Hence, the ratio of the heat transferred through a cross-section of the semicircular rod to the heat transferred through a cross-section of the straight rod in a given time is 2:π

Given:
Cross sectional area of the metal rod: A = 1 cm²= 0.0001 m²
Temperature gradient at A :
(dT/dx)_A = 5.0 °C cm^–1= 500 °Cm^-1.
Temperature gradient at B : (dT/dx)_B = 250 °C m^–1.
Heat capacity of the rod AB : C = 0.40 J°C^–1.
Thermal conductivity of the material of the rod :
K = 200 W m^–1 °C^–1.
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Rate of heat flow is dθ/dt
Hence at cross section A:



At cross section B:



Now, the rate of flow of heat throughout the rod AB is



Δθ= Q = msΔT
here, Q is the amount of heat, m is the mass of the material, s is the specific heat of the material and ΔT is the change in temperature.
And Heat Capacity is : C = ms = Q/ΔT

Substituting we get,



Hence, the rate at which temperature increases is 12.5 °C /s.

Given:
Steam temperature: T₁ = 120 °C
Length of the tube : l = 50 cm = 0.5 m
Inner radii of the tube: r = 1 cm = 0.01 m
Outer radii of the tube : R = 1.2 cm = 0.012 m
Room temperature: T₂ = 30 °C
Thermal conductivity of rubber: K = 0.15 J s^–1 m^–1 °C^–1

Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Consider an element dx at a distance of x from the center between r and R.

We will integrate this element dx to find total heat transferred from the tube.
Heat flow can be given as q = Δθ /Δt
In differential form:

Here we used negative sign because the heat flow decreases with increase in thickness dx
Also, Area of the tube formed due to element dx can be given as
A = 2πxl
Here, x is the radius of the tube due to dx and l is the length of the tube.


Now we integrate both the sides, taking temperature from tube to surrounding: T₁ to T₂ and radii from r to R






Hence, the rate of heat flow through the walls of the tube is 232.50 J/s.

Given:
Radius of the inner cylinder: r₁
Length of the cylinder= thickness of the disc: d
Radius of the disc: r₂
Temperature of inner cylinder: θ₁
Temperature of outer surface: θ₂
The thermal conductivity of the material of the disc : K

Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Consider an imaginary cylinder of radius r and thickness dr between r₁ and r₂.
We will integrate considering this imaginary cylinder to get total heat transferred.
In differential form heat flow is

Here q is the rate of heat flowing.
Negative sign indicates the decrease in rate of heat flow with increase in the thickness of the imaginary tube.
We know that area of the cylinder is:
A = 2πrd
Where r is the radius of the cylinder and d is the length of the cylinder.
Substituting we get,


Integrating both the sides we get the total rate of heat flow through the disc. Taking radius from r₁ to r₂ and temperature from θ₁ to θ₂.





Hence, the heat flowing per unit time through the disc is q.

Given data-

Length = l

Inner radius = R₁

Outer radius = R₂

Thermal conductivity = k

The corresponding diagram is shown in the fig.

a. When the flat ends are maintained at temperature T₁ and T₂(T₂> T₁)

Now, the area of cross- section through which heat is flowing is given by –

Area,

A = π (R₂² – R₁²) (1)

Let q be the heat, then

Rate of flow of heat (H)–

= (2)

H=

Where

∆T = is change in temperature between the two walls of the tube.

A= Area of cross section of the tube

K = thermal conductivity of the tube

L = length of the tube

Hence

From (1) and (2),

Rate of flow of heat -

H= k × π(R₂² – R₁²) × (T₂ - T₁)

b. When the inside of the tube is maintained at temperature T₁ and the outside is maintained at T₂.

Let’s consider a small imaginary

cylinder of radius ”r” of

differential radius “dr” as shown in fig.

Rate of flow of heat (H) –

H = (1)

Where

K = thermal conductivity of the tube

A = area of cross section

L = length of the tube

= change in temperature between the two walls of the tube.

Negative sign since “r” increases, heat decreases.

Since, the cross-section of the tube is in cylindrical form

Hence Curved Surface Area of the Cylinder,

A =2πrl

Where

r = radius of the base

l = length of the tube

From (1), substituting the value of A,

H=

⇒

=

Integrating both sides –

⇒ =

⇒

=

⇒ H =

Here the slabs are placed in such a way that their conductivities are in series as shown in the fig below

Given

Thicknesses of the slabs as L₁ and L₂

Thermal conductivities as K₁ and K₂.

Hence, their equivalent conductivity is similar to 2 resistors connected in series.

From above fig.

Given thermal conductivities of the slabs as K₁ andK₂

The of rate thermal conduction through first slab is given by -

Q_{1 =} (1)

Where

= thermal conductivity of the first slab

= area of first slab

= length of first slab

= temperature of the first slab and

T= junction temperature

Similarly the rate thermal conduction through second slab is given by –

Q₂ = (2)

Since they are connected end to end

Q_eqv = Q₁ = Q₂

Where Q_eqv is given by

Q_eqv =

Also, their area of cross section are equal ie,

A₁ = A₂ (3)

From (1) ,(2) and (3)

==

Solving for T

T = (4)

Substituting (4) in (1) and (2) –

K_{eqv =}

Given-

Conductivity of copper, K₁ = 390 W m^–1 °C^–1

Steel, k₂ = 46 W m^–1 °C^–1.

Let

length of rods = l and Area = A

Since rods are connected in series,

so the rate of flow of heat is same ie,

Q₁ = Q₂

Rate of flow of heat,

Q =

=

Since for series connected rods, Q₁ = Q₂,

⇒ =

⇒ A K₁ = A k₂

⇒ T = 10.6°C

Given

Area of cross section

A =1 cm²

=1×10^–4 m²

Thermal conductivity of aluminium, K_Al = 200 W/m°C

copper, K_Cu = 390 W/m°C

Now since these rods are connected in parallel,

So heat flowing per second

= q_Al + q_Cu

= K_Al ×A× ∆T + K_Cu ×A× ∆T

= K_Al A × (60-20) + K_Cu A × (60-20)

= 1×10^–4 m² × 40 (200+ 390)

= 2.36 W

Heat drawn in 1 second = 2.36 W

Given –

Length of each rod of 20 cm = 0.2m

Area of cross-section 0.20 cm²

= 2 × 10^-5 m²

Junction temperature = 40°C

End temperature = 80°C

The conductivities of Aluminum and copper,K_Al=200 W m^–1 °C^–1

And K_Cu = 400 W m^–1 °C^–1.

Now ,total heat drawn per second –

= Heat drawn due to copper rod + heat drawn due to Aluminium rod

= Q_Al + Q_Cu

We know, rate of heat absorption by the rod of length l, area A is given by

Q =

Where is the change in temperature

= +

⇒ Q = 2.4 J

Heat drawn in 1 minute = 2.4 × 60 = 144J

Hence, amount of heat taken out from the cold junction in one minute after the steady state is reached is 144J

For the above frame, redrawing the fig,

(1)

The equivalent resistance network becomes –

Let R_AB, R_BC, R_CD, R_DE, R_{EF and} R_BE be the equivalent resistance across each cross-section as shown in fig. below –

(2)

Resistance in terms of conductivity

R= (1)

From fig (1) and (2)and equation (1)

R_{AB =,} R_{BC =,} R_{CD =,} R_{DE =,} R_EF= and R_{BE =}

Now, lets reduce the network into equivalent network .

Since R_BC R_CD R_DE are connected in series, let R₁ be their equivalent resistance.

Then R_{1 =} R_BC+R_CD + R_DE

₌

=

Now the circuit reduces to –

Now from Kirchhoff’s current law(KCL), we know The algebraic sum of all currents entering and exiting a node must equal zero

Hence ,KCL at point E , since -

q = q₁ +q₂

Now, since R₁ and R_BE are in parallel, so total heat across R1 and R_BE will be same.

ie, q₁R₁ = q₂R_BE

⇒ =

⇒ q₂ =

Now, q = q₁ +q₂

= q₁ + q₁

Given q = 130 J, substituting above

130 =

⇒ q₁ = 60 J

From previous question,–

Now, its given bent part has k = 780 J s^–1m^–1 °C^–1 and straight part has k =390 J s^–1m^–1°C^–1for

Also Resistance equivalent circuit was as follows-

Now,

R_{AB =,} R_{BC =,} R_{CD =} R_{DE =,} R_EF= and R_{BE =}

After reducing the equivalent resistance across B and E, our circuit becomes

Where

R_{1 =} R_BC+R_CD + R_DE

₌

=

Since length is in cm and conductivity in meters, so multiply with 10^-2

R_{1 =}

Again, since R₁ and R_BE are in parallel, so total heat across R1 and R_BE will be same.

ie, q₁×R₁ = q₂ ×R_BE

⇒ =

⇒ =

The diagram is shown –

(a) Given

Thickness, l = 2 mm = 0.0002 m

Temperature inside the room = 32°C

outside = 10°C

Dimensions of wall = 1.0 m × 2.0 m

Rate of flow of heat

=

Where,

∆T = is change in temperature between the two sides of the window.

A= Area of cross section of the window

K = thermal conductivity of the window

L = length of the window

=

= 8000J/s

(b). Resistance of glass-

The equivalent circuit for the two glass panes and air becomes

Here resistance of glass R_{g =}

And of air R_{air =}

Since, these are connected in series, equivalent resistance becomes

R_{eqv =}

Thermal conductivity of window glass = 1.0 J s^–1m^–1 °C^–1

And of air, = 0.025 J s^–1m^–1 °C^–1.

Substituting values

R_eqv = ( + )

= ( + )

= 0.021

Now, rate of heat flow

=

= 380.95

Let’s take first condition –

Since, rods are connected in series, rate of heat flow remains constant.

= = constant

Where

∆T = is change in temperature between the two sides of the rods.

A= Area of cross section of the rods

K = thermal conductivity of the rods

L = length of the rod

Hence

=

Where ,

K_{A ,} K_B are thermal conductivity of rod A and B

⇒ 30 K_A = 70 K_{B ……………………………}(1)

Now, in second condition, the rods are interchanged as shown in fig below –

Let x°C be the unknown temperature of the junction AB

Now, since rate of heat flow remains constant-

=

That gives

100_B - x_B = _Ax

Substituting for _A from(1) and solving the above equation

100 = x +x

⇒

x = 30°C

(a).Let’s redraw the diagram

Since all rods are connected in series,

R_eq = R_Al + R_Cu + R_Al

Given -

Temperature of the hot end ,T₁ = 100°C

cold end ,T₂ = 0°C

Substitution for R_eq –

R_eq =

Given Thermal conductivities of aluminium =200 W m–1 °C–1

Copper = 400 W m–1 °C–1

Therefore

R_eq = + + ) =

Now rate of flow of heat,

=

= (1)

Given rate of flow of heat

From (1)

40 = 80 × 100 ×

⇒ (2)

(b) Lets redraw the diagram –

The equivalent circuit in terms of thermal resistances becomes-

R_eq = R_Al +

= ++

=

=

Now rate of flow of heat,

=

= (3)

From (2)

Substituting in (3)

=

= 75 W

(c) Lets redraw the diagram-

Since the thermal resistors are connected in parallel.

= + +

⇒ =

⇒ = (_Al + _Al +_Cu ) = (200+200+400)

⇒ = *(800)

⇒ =

From (2)

⇒ =

⇒ = 400W

Let’s redraw the diagram

Let the temperature at junction B be T.

Let Q_A, Q_C and Q_B be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.

From fig.

At point B

Q_A = Q_C + Q_B

Now, rate of flow of heat is given by –

= +

⇒ = +

⇒ = +

⇒ T =

Given thermal conductivity of the respective rods as follows-

K_A = K_C = K₀
K_B = K_D = 2K₀
K_E = 3K₀, K_F = 4K₀
K₉= 5K₀

Also, length of each rod is l

At steady state, temperature at the ends of rod F will be same.

Let T be the temperature of rod F

(a)

Rate of heat flow through rod A + rod C

= Rate of heat flow through rod B + rod D

⇒ + = +

Substituting the values in terms of k₀ –

⇒ + = +

⇒ 2k₀ (T₁-T) = 2 2 k₀ (T–T₂)

⇒ T =

(b) To find the rate of flow of heat from rod G, which is at Temperature T₂

Looking into the above diagram, we can say that it forms a balanced Wheatstone bridge.
Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.

Hence we can remove the F for simplification

From above diagram, we can see that R_A and R_B are connected in series.

⇒ R_AB = R_A + R_B

And R_C and R_D are connected in series

⇒ R_CD = R_C + R_D

Then, R_AB and R_CD are connected in parallel

Now ,

R_A = , R_B = , R_C = , R_D =

Since R_{A ,} R_B are connected in series

R_AB = and R_CD =

Since R_AB R_CD are in parallel

=

=

=

Now, rate of flow of heat from the source rod

q = =

=

Hence, rate of flow of heat from the source rod is given by

q =

Let’s redraw the diagram

From the above diagram we can say that ΔABE is similar to ΔACD.

By the property of similar triangles-

=

⇒ x = r₁ + (r₂ – r₁ )

Lets assume-

a =

⇒ r = ax + r₁ (1)

Thermal resistance is given by –

dR =

Now area

dR =

dR =

=

Solving above integral

R =

⇒ R =

–

Rate of heat flow =

q = Kπr₁r₂

Given-
Length of the rod, l = 20 cm = 0.2 m
Area of cross section of the rod, A = 1.0 cm² = 1.0×10^-4 m²

Thermal conductivity of the rod, k = 200 W m-1°C-1

Also, the temperature of one end is maintained at 0°C and that of the other end is varied from 0°C to 60°C in 10 minutes.

Hence rate of increase of the temperature at one end is 0.1°C per second.

⇒ = = 0.1°C/s

Now, rate of flow of heat –

=

= +...........+ (

= )

We know Arithmetic Progressions –

For a series given by

a+2a+...................na

Sum of n terms is given by

Sum= , where a is the first term, d is the difference between first and second term and n is the number of terms.

Substituting the values, for 10 minutes –

Here a= 0,1. d = 0.1 and n = 60 for 10mins,

=

= 1800J

Hence , the total heat transmitted through the rod in these 10 minutes is 1800J

Let’s redraw the circuit –

Let-
Radius of the inner sphere =a
Radius of the outer sphere = b

Given –

a = r₁ = 5cm = 0.05m

b = r₂ = 20cm = 0.2m

θ₁ = T₁ = 50°C

θ₂ = T₂ = 10°

Consider an imaginary shell of radii r and thickness dr.

Area,

Now, rate of flow of heat –

=-

= is change in temperature.

A= Area of cross section of the tube

K = thermal conductivity of the tube

d_r = change in length

Here, the negative sign is for decrease in temperature with increase in radius.

q = -

Taking integral on both sides –

=

Solving above integral –

q= = k( ) = 100 (given)

Substituting the values,

K = = 2.8 = 3 W m-1°C-1

Given length of metal rod = L

specific heat capacity of water = s

Mass of water =m

Rate of transfer of heat –

=

= is change in temperature.

A= Area of cross section of the tube

K = thermal conductivity of the tube

L= length

In time ∆t, the heat transfer from the rod will be given by

∆Q= (1)

Now, heat loss by water at temperature T₁ is equal to the heat gain by water at temperature T₂

So, heat loss by water at temperature T₁ in time ∆t is -

∆Q=ms(T₁-T₁’) (2)

Where

m = mass of water

S = specific heat of water

From (1) and (2)

⇒ ms(T₁-T₁’) =

T₁’=

This is the fall in temperature of water at temperature T₁.

Similarly, rise in temperature of water at temperature T₂

T₂’=

Finally, change in temperature is given by-

(T₁’-T₂’)= –

⇒ {(T₁’-T₂’)- (T₁-T₂)} = -

⇒ = -

Where is the rate of change of temperature difference.

Taking integral on both sides-

= dt

⇒ t =

Hence the time taken for the difference between the temperatures in the vessels to become half of the original value is

t =

Given-

Masses of body = m₁ and m₂

Specific heat capacities = s₁ and s₂

Rod of length= ℓ,

Cross-sectional area = A

Thermal conductivity = K

Rate of transfer of heat from the rod is given by –

=(1)

Where,= temperature of first and second body.

A= Area of cross section of the

K = thermal conductivity of the

L= length

Heat transfer from the rod in time ∆t –

∆Q= (2)

Heat loss by the body at temperature T₂ is equal to the heat gain by the body at temperature T₁.

Heat loss by the body at temperature T₂ in time ∆t is –

∆Q=m₂s₂ (T₂′-T₂) (3)

From (1) and (2)

m₂s₂ (T₂′-T₂) =

⇒ T₂’=T₂- ∆t

This is the fall in the temperature of the body at temperature T₂.

Similarly, rise in temperature of water at temperature T₁ is –

T₁’=T₁+ t

Change in the temperature

(T₂′-T₁′)

= T₂- ∆t - T₁- t

⇒ {(T₂′-T₁′)-(T₂-T₁)}=- ∆t - t

⇒ = -

Where is the rate of change of temperature difference

⇒∆T= )

Integrating both the sides -

=

⇒ ln |T₂-T₁| = -

Taking the anti-log

⇒ (T₂-T₁)=e^-λt

Where

λ=

Given,

In time dt, heat transfer through the bottom of the cylinder is given by-

= (1)

In case of monoatomic gas, pressure remains constant.

Hence the heat content at constant pressure(enthalpy) is given by

dQ=nC_pdT (2)

where,

dQ=change in heat

n = number of molecules

dT = change in temperature

C_p = amount of heat required to raise the temperature of a substance of 1Kg mass by one degree Celsius at constant pressure.

Comparing above equations-

=

For a monoatomic gas, C_p=52 R

⇒ =

⇒ =

⇒ =-

Integrating both the sides, we get

=

=-

⇒ ln( =-

Taking antilog

⇒ =

⇒ T =

Rewriting

⇒ T- = ) (1)

Now, we know the gas equation given by

=

Substituting in (1)

= T- = )

Solving for the length/distance,

l = )

Given-
Area of the body, A = 1.6 m²

Temperature of the body, T = 310 K
We know from Stefan-Boltzmann law, we have-

= σAT⁴

Where,

A is the area of the body

σ is the Stefan-Boltzmann constant

σ = 6 × 10⁻⁸ W m⁻² K⁻⁴.

Therefore,

Energy radiated per second = 1.6 × 6 × 10⁻⁸ × (310)⁴
= 886.58

≈887 J

Given

Area of the body, A = 12 × 10⁻⁴ m²

Temperature of the body, T =20°C

= (273 + 20) K

= 293 K

Emissivity of the surface, e = 0.80

Stefan-Boltzmann constant σ = 6.0 × 10⁻⁸ W m⁻² K⁻⁴.

Now

Rate of emission of heat is given by-

R = AeσT⁴

Where

A = Area of the surface

e = Emissivity of the surface

σ = Stefan-Boltzmann constant

And T = temperature

Substituting the values -

⇒ R = 12 × 10⁻⁴ × 0.80 × 6.0 × 10⁻⁸ × (293)⁴

⇒ R = 0.42 J