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)