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].

AP Physics B Multiple Choice Practice Questions on Temperature and Heat

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives”

– Albert Einstein

As promised in the last post dated 6^th February 2009, I give below a few multiple choice practice questions on temperature and heat relevant to AP Physics B:

(1) A soft iron rod has its length increased by 0.3% on increasing its temperature by ∆T. The percentage increase in the area of a circular hole in a sheet of the same material (soft iron) on increasing its temperature by ∆T will be

(a) 0.3%

(b) – 0.3%

(c) 0%

(d) 0.6%

(e) –0.6%

The radius of the hole will increase by 0.3% and hence the area of the hole will increase by 0.6%.

[The area A = πR² where R is the radius. Therefore, ∆A/A =2 ∆R/R. The fractional change in area is equal to twice the fractional change in radius. Therefore, the percentage change in area is equal to twice the percentage change in radius].

(2) The temperature of oil in a vessel is measured using a Celsius scale thermometer as well as a Fahrenheit scale thermometer. The readings are respectively tº C and 2tº F. The temperature of the oil is

(a) 32º C

(b) 100º C

(c) 120º C

(d) 142º C

(e) 160º C

We have t_C/100 = (t_F – 32) /180.

[This can be remembered as t_C = (5/9)(t_F – 32)]

Here t_C = t and t_F = 2t so that t/100 = (2t– 32) /180.

Or, 9t = 10t– 160 from which t = 160.

Therefore, the temperature of oil is 160º C.

(3) A copper rod and a steel rod are to have lengths Lc and L_s such that the difference between their lengths is the same at all ambient temperatures. If the coefficients of linear expansion of copper and steel are respectively α_c and α_s the lengths are related to the coefficients of linear expansion as

(a) Lc/L_s = α_c/α_s

(b) Lc – L_s = α_c – α_s

(c) Lc/L_s = α_s/α_c

(d) Lc/L_s = (α_c/α_s)^1/2

(e) Lc/L_s = (α_c/α_s)^–1/2

The changes in length must be equal on heating (or cooling) so that we have

Lc α_c ∆T = Ls α_s ∆T

This gives Lc/L_s = α_s/α_c

(4) Four identical brass rods AB, BC, CD and DA are joined to form a square (fig). If the diagonally opposite corners A and C are kept in melting ice and boiling water respectively what is the temperature difference between the corners B and D?

(a) 50º C

(b) 100º C

(c) 25º C

(d) 75º C

(e) 0º C

The corners B and D will be at the same temperature so that the difference will be zero.

(5) Thermal conductivity of metal A is three times the thermal conductivity of metal B. Two rods of the same dimensions made of metals A and B are joined end to end (fig.). The free end of A is maintained at 0º C and that of of B is maintained at 100º C. The temperature of the junction between A and B in the steady state will be

(a) 25º C

(b) 33.3º C

(c) 66.6º C

(d) 75º C

(e) 90º C

In the steady state the quantity of heat flowing through B per second will be equal to that flowing through A per second. If the temperature of the junction between A and B is tº C we have

kA(100 – t)/L = 3kA(t – 0)/L where A is the area of cross section, L is the length, k is the thermal conductivity of B and therefore 3k is the thermal conductivity of A.

Therefore 100 – t = 3t so that t = 25º C

[Since the rods are of identical dimensions and the thermal conductivity of A is three times that of B, you can easily argue that the temperature drop in A is only one-third of the temperature drop in B and arrive at the answer].

(6) A cup of coffee cools from 65º C to 63º C in one minute in a room of temperature 24º C. To cool from 58º C to 50º C the same cup of coffee will take nearly

(a) 2 min.

(b) 2.4 min.

(c) 3.8 min.

(d) 4.5 min

(e) 5.3 min

Since the rate of cooling is directly proportional to the excess of temperature (in accordance with Newton’s law of cooling), we have

S(65 – 63)/1 α (64 – 24) and

S(58 – 50)/t α (54 – 24)

where S is the heat capacity (thermal capacity) of the cup of coffee (which is the heat required to raise the temperature through 1K) and t is the time required to cool from 58º C to 50º C. Note that we have used the average temperatures (64º C and 54º C) of the cup of coffee to find the mean excess temperatures in the two cases.

On dividing, 2t/8 = 40/30 from which t = 5.3 min (nearly)

Temperature and Heat for AP Physics B- Equations to be Remembered

Here are the essential things you need to remember for solving multiple choice questions involving heat and temperature:

(1) Temperature t_C in Celsius scale can be converted to temperature t_F in Fahrenheit scale using the relation

t_C/100 = (t_F – 32) /180

Remember that the ice point and steam point of water are 32º F and 212º F respectively in the Fahrenheit scale. These temperatures in the Celsius scale are 0º C and 100º C respectively so that a temperature difference (∆t_F) of 180º in the Fahrenheit scale is equal to a temperature difference (∆t_C) of 100º in the Celsius scale. The significance of the numbers 100, 32 and 180 will be clear to you now.

(2) Temperature t_C in Celsius scale can be converted to temperature t_K in Kelvin scale (absolute scale) using the relation

t_C/100 = (t_K – 273.15) /100

so that t_C = t_K – 273.15 or, t_K = t_C + 273.15.

Often the above relation is written as t_K = t_C + 273 very nearly.

The above equation follows from the fact that the ice point and steam point of water are 273.15 K and 373.15 K respectively in the Kelvin scale.

A temperature difference of 1º in the Kelvin scale is equal to a temperature difference of 1º in the Celsius scale.

(3) The increase in length (∆ℓ) of a solid on raising its temperature by ∆T is given by

∆ℓ = α ℓ_o ∆T where α is the coefficient of linear expansion (linear expansivity) and ℓ_o is the original length.

(4) The increase in area (∆A) of a solid on raising its temperature by ∆T is given by

∆A = β A_o ∆T where β is the coefficient of area expansion (area expansivity) and A_o is the original area.

In the case of isotropic homogeneous solids β = 2α

(5) The increase in volume (∆V) of a solid on raising its temperature by ∆T is given by

∆V = γV_o ∆T where γ is the coefficient of volume expansion (volume expansivity) and V_o is the original volume.

In the case of isotropic homogeneous solids γ = 3α

(6) Specific heat capacity (specific heat) C of a substance is the quantity of heat absorbed or rejected by 1 kg of the substance to change the temperature by 1 K.

Therefore, the heat involved in changing the temperature of m kg of a substance through ∆T K (or ∆T º C) is mC ∆T where C is the specific heat of the substance.

(7) Molar specific heat of a substance is the quantity of heat absorbed or rejected by 1 mole of the substance to change the temperature by 1 K.

You know that in the case of gases there are two specific heats viz., specific heat at constant volume and specific heat at constant pressure. Similarly there are two molar specific heats viz., molar specific heat at constant volume and molar specific heat at constant pressure. In this context you will find a useful post here.

(8) Heat transfer takes place by three processes viz., conduction, convection and Radiation.

The quantity of heat Q conducted in a time t through a uniform rod of length L and area of cross section A when the ends of the rod are maintained at a temperature difference ∆T is given by

Q = KA (∆T/L) t where K is the thermal conductivity of the material of the rod

[The temperature gradient ∆T/L is often written as ∆T/∆x, replacing L by ∆x, which will be more appropriate in the case of heat conduction through slabs of thickness ∆x].

The rate of transfer of heat by conduction (time rate), H is given by

H = Q/t = KA ∆T/L

According to Stefan’s law the energy (E) radiated per second per unit surface area of a perfectly black body is directly proportional to the 4^th power of the absolute temperature of the body:

E α T⁴

Or, E = σT⁴ where σ is Stefan’s constant.

So if the temperature is doubled, the energy radiated from the body will become 16 times.

Newton’s law of cooling says that the rate of cooling (rate of loss of heat) of a body is directly proportional to the excess of temperature of the body over the surroundings:

dQ/dt α (T₂ – T₁) where dQ is the heat lost in a time dt when the temperatures of the body and the surroundings are respectively T₂ and T₁.

Note that the loss of heat mentioned in Newton’s law of cooling is due to all the three mechanisms viz., conduction, convection and radiation. Further, the excess temperature (T₂ – T₁) should be small.

In the next post we will discuss questions in this section. Meanwhile find a useful post here.