Thermal Properties Of Matter

First of all understand the bimetallic strip, when two strips of equal length but of different materials, means different coefficient of linear expansion, join together then this is known as bimetallic strip. This strip has the property of bending on heating due to unequal linear expansion of two metals.

Both strips of Al and steel are fixed together initially in bimetallic strip. When both are heated then expansion in steel will be smaller than aluminium.

So, Al strip will be convex side and steel on concave side.

Therefore option (d) is correct.

As we know moment of inertia of a rod about its perpendicular bisector is so from this formula we can say easily that if L increases then I also increases, so on heating a uniform metallic rod its length will increase so moment of inertia of rod increased from I₁ to I₂ an obviously I₂ > I₁ or .

Now also we will apply conservation of angular momentum i.e.

I₁ω₁= I₂ω₂ or so from here ω₁>ω₂ which means angular speed decreases.

Therefore option (b) is correct.

As we know the identity by which temperature on one scale can be converted into another scale i.e.

where LFP & UFP stands for Lower fixed point & upper fixed point respectively.

Now from above graph we can say Lower fixed point is

and Upper fixed point

Hence applying the above formula for A and B

i.e.

Therefore option (b) is correct.

As we know that buoyant force on a body of volume V and density of ρ, when immersed in liquid is F_B.

So at 0℃ assume the volume be V and density be ρ₀ then buoyant force F_0℃and at 4℃ F_4℃ also density i.e. ρ₄ will be maximum all over the range of temperature so

Now ratio of buoyant force

So

Therefore option (a) is correct.

As we know time period of a pendulum is given by so from this we can say

Now also we know that as the temperature increased the length L increases due to linear expansion i.e. .

So from above we can say on increasing temperature, its effective length increases hence T also increases.

Therefore option (a) is correct.

We know that vibration of molecules about their mean position increases as the temperature increases or the body get heated.

Hence kinetic energy associated with random motion of molecule increases.

Therefore option (a) is correct.

As we know the volume of sphere

Also given in question that coefficient of linear expansion is also coefficient of cubical expansion

So

Option (d) is matching hence correct option is (d).

loss of heat on cooling is directly proportional to the following factors

(i) surface area exposed to surrounding

(ii) temperature difference between body and surrounding

(iii) material of object

Also we know that surface area of sphere is minimum and of circular plate is maximum.

So option (c) s correct and rests are wrong.

In this question we will apply Zeroth law of thermodynamics which state if two thermodynamics systems are each in thermal equilibrium with third one, then they are in thermal equilibrium with each other.

So look option (a) here T_x=T_y but T_x ≠ T_z hence T_y ≠ T_z which means Y and Z never in thermal equilibrium so option (a) is wrong.

In option (b) T_x=T_y but T_x ≠ T_z hence T_y ≠ T_z so this option is correct.

In option (c) T_x≠ T_y also T_x ≠ T_z hence T_y ≠ T_z so this option is also wrong.

And at last in option (d) T_x≠ T_y and T_x ≠ T_z hence T_y may be equal T_z so option (d) is correct.

As we know, material whose surface area less will get heated before than the more surface area.

Hence options (b) and (c) are correct and rest options are wrong.

As we know whenever any substance get heated continuously but its temperature not changing, then obviously the state changes i.e. solid to liquid or liquid to gas or similarly.

Now look carefully the above graph, we are seeing that part AB (0℃) and CD (100℃) represents the conversion of solid into liquid and liquid into gas respectively. That means part AB represents ice and water both till B, and part CD represents water and steam.

So from above we can say that option (b) is wrong and also option (c) is wrong because at D all waters get converted into steam. Hence options (a) & (d) are correct.

Lost of heat is directly proportional to the temperature difference with surrounding and body. So as milk cool, temperature difference decreases so rate of cooling decreases with time so from here we can say option (a) is wrong.

According to Newton’s law of cooling the heat of milk falls exponentially so option (b) is correct.

While cooling very small amount of heat also flows from surrounding to milk as compared to heat lost by milk to surrounding so option (c) is also correct.

By conduction, convection and radiation, when hot milk spread on table it transfer heat to surroundings. Hence option (d) is correct.

As we know diathermic walls allows to conducts heat through it while adiabatic walls doesn’t allow. Hence bulb of a thermometer made of diathermic walls.

As we know, so from 1^st observation we can say that linear expansion of this rod will be

. Now this linear expansion will be same for all observations. Now we will check for rest of observations, so for 2^nd observation

but in question

which is not equal to our calculated value hence this observation is wrong.

Similarly for 3^rd observation

but given hence this observation is also wrong. Now last observation

which is correct.

The law of conduction of heat depends on the conductivity of the material.

Here, “H” is the rate of heat transfer, “K” is the thermal conductivity constant, A is the area of cross-section, T₂-T₁ is the temperature difference.

Metals have higher conductivity than wood. This means when we touch hot metals or wood, the heat from the metals flow after towards our hands as compared to that of wood. So, they appear hotter. The same theory goes when the two bars are cooled. Here, the heat from our body has a least resistive path when we touch metals as they have higher conductivity. So, heat is dissipated faster in metals than in wood and we feel them cooler.

The relation between Celsius and Fahrenheit is as follows:

Here, temperature on Celsius is same as Fahrenheit. This means

Here, x is any variable.

So, at -40°C we have the same temperature in Fahrenheit scale i.e. -40°F.

The use of copper with steel in utensils has to do with their different conductivities. The heat transfer depends on the conductivity of the material and is given as:

Here, “H” is the rate of heat transfer, “K” is the thermal conductivity constant, A is the area of cross-section, T₂-T₁ is the temperature difference.

Copper has higher conductivity and so is heated immediately. Steel on the other hand has lower conductivity. Because of this, the transfer of heat at the intersection of steel and copper is a gradual process. This ensures uniform heating of food.

The moment of inertia of a uniform rod (along perpendicular bisector) is given by:

————— (1)

where, I is the inertia, M is the mass of the rod and l is its length.

The thermal expansion ΔL is given by:

————— (2)

Here, L’ is the new length, L was the original length, 𝛂 is the thermal expansion coefficient and ΔT is the temperature difference.

————— (3)

Substituting (3) in (1)

We know the change in length, ΔL is very small. So, we can neglect square terms like (𝛂ΔT)². Thus, we get,

Now,

Therefore, the increase in moment of inertia is 2I𝛂ΔT.

The process of making an ice ball deals with interconversion of the states of water.

While making an ice ball, we apply pressure on the crushed ice. The temperature is 0°C and the atmospheric pressure is 1 atm. Increasing this pressure shift the state from solid to liquid as indicated by the circle. Due to this, a part of the crushed ice melts and fills any gaps present in between ice flakes. When we release pressure, the water again solidifies to ice thus making the ice ball.

Here, it’s given

Mass of water, m=100g

Cooled temperature, T₁=-10°C

Specific heat capacity of water, S_w= 1 Cal/g/°C

Latent heat of fusion, L_w=80 Cal/g

Let the mass which would freeze be m’.

Ice forms at 0°C.

So, the temperature of the resultant mixture will be T₂ = 0°C

We know,

Hence, the temperature of the resultant mixture is 0°C and the mass which would freeze is 12.5g.

The rate of heat energy loss is directly proportional to the difference in temperature. In the first scenario, mixing the cold water with the hot water overall brings the water to an intermediate temperature which isn’t too hot. The difference between this intermediate temperature and the surrounding temperature is less. Hence the heat loss would be less. In the second scenario, the temperature difference between the hot water and the surrounding is considerable and thus, the heat loss will occur rapidly. This will cool the water down.

So, the first option will keep the water warmer.

Let the length of iron strip be L_I and that of brass be L_B. It is given that the length of the scale should be 10 cm. So,

————— (1)

Both the strips expand at different rates due to different expansion coefficient. But the net expansion in the strips needs to be equal to maintain 10cm length. This means,

————— (2)

————— (3)

————— (4)

(as, α_iron 1.2 X 10^-5/ K and α_brass 1.8 X 10^-5 /K)

—————(5)

Substituting (5) in (1)

The length of the brass rod should be 20 cm and that of the iron rod be 30 cm to maintain 10 cm length at all temperatures.

Here, we can have a bimetallic vessel with iron on the outside and brass on the inside as brass expands more than iron.

The change in volume of both the metals should be same.

Now,

So, the volume of the brass vessel should be 144.9 cc and that of iron vessel should be 244.9 cc.

Given,

Coefficient of linear expansion, α = 1.7× 10^-5 /°C

Therefore, Coefficient of volumetric expansion(𝛾)

𝛾 = 3α = 1.7× 10^-5 /°C = 5.1× 10^-5 /°C

Bulk modulus, B= 140 × 10 ⁹ N/m².

Body temperature, T1 = 37°C

Temperature of tea, T2 = 57 °C

Temperature difference, ΔT = 57 - 37 = 20 °C

We know,

Now, stress,

Atmospheric pressure is 10⁵ N/m².

So, the stress is about 10³ times atmospheric pressure.

Here, we apply Pythagoras theorem to find x in terms of L.

From the figure, it is clear that

As, ΔL² is a very small quantity, we can neglect it.

(here, we multiply and divide 2 in the numerator and denominator)

We know,

Here,

α_steel = 1.2× 10^-5 /°C and ΔT = 20 °C. Therefore,

So, the displacement is about 11 cm.

The length of the rod is L0. And the coefficient of linear expansion is α. Therefore, temperature (𝜃) at any length x is:

Where 𝜃₁ and 𝜃₂ are temperatures at the two ends.

We know,

For a small change dx in length dx’ for temperature 𝜃, we have

Integrating the equation,

Now, (original length) and (NEW LENGTH)

Therefore,

(ANS)

Given,

The radius of the weapon, r=0.5 m

Therefore, Surface area, A=4𝜋r² = 𝜋

Temperature after detonation, T=10⁶ K

Stefan’s constant, σ = 5.67 × 10^-8 W/ m²K⁴

(a) Therefore, power radiated (P) is given by

(ANS)

(b) Here, 10% of the energy(E) is to evaporate water which is at 30°C. This means the energy heats the water up to 100°C and then evaporates.

So, 0.1E = msΔT +mL_v

Here, the first term is due to the contribution of energy in heating the liquid to 100°C and the second term is due to vaporization and m is the mass of the water used.

Energy,

Here, t is the time which is 1 sec.

Given, Heat capacity, S_w = 4186.0 J/ kgK

Latent heat of vaporization, L_v = 22.6 X 10⁵ J / kg

Temperature difference, ΔT = 100°C - 30°C = 70°C

So, 7x10⁹ kg of water would be evaporated.

(c) Here, energy,

Therefore, momentum,

Now, momentum at a distance 1km (per unit time) is

Therefore, the momentum imparted at 1km is 47.7 N/m²