Multiple Choice Practice Questions on Heat Transfer and Thermal Expansion for AP Physics B

“Genius is one percent inspiration and ninety nine percent perspiration”
– Thomas A. Edison

Today we will discuss some practice questions (MCQ) on heat transfer and thermal expansion. You may click here to obtain the essential points you need to note in this section.
Here are the questions:
(1) When water is heated from 0º C to 20º C its volume
(a) goes on increasing
(b) goes on decreasing
(c) remains constant up to 15º C and then increases
(d) first decreases and then increases
(e) remains constant up to 4º C and then increases
This question is meant just for checking your knowledge of the behaviour of water. Water has maximum density at nearly 4º C and hence the correct option is (d).
(2) 5 g of ice at 0º C is mixed with 10 g of water at 10º C. The temperature of the mixture is
(a) 0º C
(b) 2º C
(c) 2.5º C
(d) 5º C
(e) 7.5º C
In order to melt 5 g of ice (into water) without change of temperature 400 calories of heat are required since the latent heat of fusion for ice-water change is nearly 80 calories per gram. The heat that is released by 10 g of warm water at 10º C on cooling to 0º C is 100 calories only since the specific heat of water is 1 calorie per gram per Kelvin.
So the warm water can melt just a quarter of the amount of ice and the mixture will remain at 0º C [Option (a)].

(3) Equal masses of three liquids of specific heats C₁, C₂ and C₃ at temperatures t₁, t₂ and t₃ respectively are mixed. If there is no change of state, the temperature of the mixture is

(a) (t₁+ t₂ + t₃)/3

(b) (C₁t₁+ C₂t₂ + C₃t₃)/[3(C₁+ C₂ + C₃)]

(c) (C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(d) 3(C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(e) 3(t₁+ t₂ + t₃)

If the mass of each liquid is m, the total amount of heat (H) initially is given by

H = m(C₁t₁+ C₂t₂ + C₃t₃)

After mixing the same amount of heat is available. If the common temperature is t, we have

H = mt(C₁+ C₂ + C₃)

From the above equations, t = (C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(4) The amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is (universal gas constant = R)
(a) 2 R
(b) 3 R
(c) 5 R
(d) 5R/2
(e) 7R/2
The molar specific heat of a mono atomic ideal gas at constant pressure (cp) is 5R/2 where R is the universal gas constant.
[The molar specific heat of a mono atomic ideal gas at constant volume (cp) is 3R/2 and in accordance with Meyer’s relation, cp = cv + R].
Therefore, the amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is 1×(5R/2) ×2 = 5R.

(5) Two identical rectangular strips, one of copper and the other of steel, are riveted as shown to form a bi-metal strip. On heating, the bi-metal strip will

(a) get twisted

(b) remain straight

(c) bend with steel on the convex side

(d) bend with steel on the concave side

(e) contract

On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side.

[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].

(6) Four cylindrical rods of different radii and lengths are used to connect two heat reservoirs at fixed temperatures t₁ and t₂ respectively. From the following pick out the rod which will conduct the maximum quantity of heat:

(a) Radius 1 cm, length 1 m

(b) Radius 1 cm, length 2 m

(c) Radius 2 cm, length 4 m

(d) Radius 3 cm, length 8 m

(e) Radius 0.5 cm, length 0.5 m

The quantity of heat conducted is directly proportional to the area of cross section and inversely proportional to the length of the rod (when the same temperature difference exists between the ends).

[Remember that the quantity of heat Q = KAdθ/dx where K is the thermal conductivity, A is the area of cross section and dθ/dx is the temperature gradient].

When you compare rod (a) with rod (b) you find that rod (a) can conduct better since its length is less than that of rod (b).

Rod (a) and rod (c) conduct equally since the cross section area as well as the length of rod (c) is 4 times that of rod (a).

Rod (d) is better than rod (a) since its cross section area is 9 times that of rod (a) while its length is only 8 times that of rod (a).

Rod (e) is worse than rod (a) since its cross section area is a quarter of that of rod (a) while its length is half that of rod (a).

Therefore, the rod which will conduct the maximum quantity of heat is (d).

[It is enough to compare the ratio of the area to length. The area is directly proportional to the square of the radius. Since the unit of radius is centimetre and that of length is metre in all cases, you can blindly compare the values of 1²/1, 1²/2, 2²/4, 3²/8 and (0.5)²/(0.5). The highest value is 3²/8].