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Name: ANSWER Date:

Class & Class No: 3 ( )

1. In calibrating an unmarked liquid-in-glass thermometer, it is found that the liquid thread is 3 cm long in

melting ice and 21 cm long in boiling water. When it is put into a glass of water, the thread is 9 cm long.

What is the temperature of the water?

Increase in length by 1 o C = (21 – 3)/(100 – 0)

= 0.18 cm o C –1

(9 – 3)/(T – 0) = 0.18

T = 33.3 o C

2. A 1 kW heater takes 138 s to heat a tank of oil from 25 o C to 50 o C. If the mass of oil is 1.8 kg, calculate

the heat capacity and the specific heat capacity of the oil.

Q = m c (T 2 – T 1)

1 000 (138) = 1.8 c (50 – 25)

c = 3 070 J kg –1 o C –1

3. Tom carried out an experiment to determine the specific heat capacity of a metal block using an

immersion heater and joulemeter. He obtained the following results:

mass of metal block = 1 kg

initial joulemeter reading = 12 345 J final joulemeter reading = 16 123 J

initial temperature of metal block = 25 o C final temperature of metal block = 29 o C

Determine the specific heat capacity of the metal from the above data.

Q = m c (T 2 – T 1)

(16 123 – 12 345) = 1 c (29 – 25)

c = 944.5 J kg –1 o C –1

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4. A student carried out an experiment to find the specific heat capacity of water by heating 200 g of

water with an 20 W immersion heater for 300 s. The graph below shows the variation of the

temperature of water with time.

t / s

T / Co

32

2526

30 300

graph A

graph B

(a) (i) Calculate the energy supplied by the heater from t = 30 s to 300 s.

(ii) Using the result obtained from (a)(i), determine the specific heat capacity of water.

(i) E = P t

= 20 (270)

= 5 400 J

(ii) Q = m c (T 2 – T 1)

5 400 = 0.2 c (32 – 26)

c = 4 500 J kg –1 o C –1

(b) Sketch, on the graph given, the variation of the temperature of water if

(i) the mass of water is doubled (labelled as graph A); and (see the figure above)

(ii) the power of the heater is doubled (labelled as graph B). (see the figure above)

5. Liquid X at 10 o C is mixed with an equal mass of liquid Y at 50 o C. If the ratio of the heat capacity of X

to Y is 1 : 3, find the final temperature of the mixture.

Energy lost by liquid Y = Energy gained by the liquid X

m c Y (50 – T) = m c X (T – 10)

m (3c) (50 – T) = m c (T – 10)

T = 40 o C