2009 JC1 H2 Physics
FA3: Thermal Physics
Instructions:
26 May 2009
- Answer all questions in the spaces provided using a dark blue or black pen.
Duration: 25 min
- All workings must be shown clearly.
- Marks will be deducted for inappropriate significant figures, incorrect units or use of g = 10 m s-2.
1
(a) State what is meant by the specific heat capacity of a material.
[1]
The specific heat capacity of a material is defined as the amount of thermal energy
required to cause a unit rise in the temperature of a unit mass of the material.
(b) The graph in Fig. 1.1 shows the variation with time t of temperature change  for
1.0 kg of substance, initially solid at room temperature.
8.0
 / K
7.0
6.0
5.0
4.0
3.0
2.0
1.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10
t / min
Fig 1.1
The substance is heated at a uniform rate of 2000 J min-1.
(i) State the time at which all the substance has just changed to liquid.
[1]
From 0 to 4.0 min: the temperature of the substance is increasing, thus the
substance is still in its solid state.
From 4.0 to 7.0 min, the temperature of substance remains constant, indicating
that it is melting. Thus the substance will consist of solid and liquid.
From 7.0 min onwards, the temperature the substance starts to increase again,
indicating the end of the melting process where the substance is comprised of
just liquid.
time = 7.0 min
(ii) With reference to Fig 1.1, deduce with reasons if the specific heat capacity of the
substance is lower when it is in liquid or solid state.
[2]
Q  Pt  mc
P
 
t
mc
P
. With the same power
mc
and mass of substance, the gradient of Δθ against t is inversely proportional to
Thus the gradient of Δθ against t reflect the value of
1
specific heat capacity. [1] Since the rate of change of temperature is higher when
the substance is comprised of liquid, the specific heat capacity of the liquid is
lower than that of the solid. [1]
(iii) Calculate the specific latent heat of fusion of the substance.
Q  Pt  mLf
2000  (7.0 - 4.0) = (1.0) Lf
Lf = 6.00 × 103 J kg-1
2
[1]
[1]
(a) State what is meant by internal energy of an ideal gas.
[1]
Internal energy of an ideal gas is the sum of a random distribution of microscopic
kinetic energy of all the molecules in the system (since microscopic potential energy
is zero for an ideal gas).
(b) A fixed mass of gas in a heat pump undergoes a cycle of changes of pressure,
volume and temperature as illustrated in Fig. 2.1. The gas is assumed to be ideal.
Fig 2.1
The table below shows the increase in internal energy which takes place during each
of the changes A to B, B to C and C to D. It also shows that in both of sections A to
B and C to D, no heat is supplied to the gas.
Increase in internal Heat supplied
energy / J
gas / J
to Workdone on gas
/J
A to B
1200
0
(1) + 1200
B to C
-1350
(3) -1350
(2) 0
C to D
-600
0
(1) -600
D to A
(4) + 750
(5) + 750
(2) 0
A to B to C to D to A
(4) 0
(6) - 600
(6) + 600
(i) Using the first law of thermodynamics and necessary data from the graph,
complete the table. You will find it helpful to proceed in the following order.
[7]
1. Work done on gas for A to B and C to D
2
ΔU = Q + W
For A to B: 1200 = 0 + W AB
WAB = + 1200 J
For C to D: - 600 = 0 + W CD
WCD = - 600 J
2. Work done on gas for B to C and D to A
Since there is no change in the volume during B to C and D to A, the work
done on the gas during B to C and D to A is both zero.
3. Heat supplied to gas from B to C
ΔU = Q + W
For B to C: - 1350 = QBC + 0
QBC = - 1350 J
4. Increase in internal energy for D to A
To get the ΔUDA, the value of ΔUABcDA needs to be found first. Since the
ideal gas undergoes a cyclic process, the starting and ending temperature is
the same, i.e. ΔT = 0. With the internal energy of an ideal gas proportional to
its temperature, ΔUABcDA = 0.
ΔUABcDA = 0
ΔUAB + ΔUBC + ΔUcD + ΔUDA = 0
(+ 1200) + (- 1350) + (- 600) + ΔUDA = 0
ΔUDA = + 750 J
5. Heat supplied to gas from D to A
ΔU = Q + W
+ 750 = QDA + 0
QDA = + 750 J
6. net Q
= QABcDA
= Q A  B + Q B  C + Q c D + Q D  A
= (0) + (- 1350) + 0 + (+ 750)
= - 600 J
net W
= WABcDA
= WAB + W BC + W cD + W DA
= (+ 1200) + (0) + (- 600) + (0)
= + 600 J
(ii) Calculate P, the coefficient of performance of the heat pump, given that
Heat delivered by gas (during change B to C)
P
Net work done on gas
P
[1]
1350
 2.25
600
3
Bonus Question
Using the Kinetic Theory of Gas, outline how molecular movement causes the pressure
exerted by a gas?
[2]
When the gas molecules collide with the walls of its container, the gas molecules undergoes
a change in momentum over time. The change in momentum is caused by the force exerted
on the gas molecules by the wall. At the same time, during the interaction of the gas
molecules and the wall, the gas molecules will exert a force back onto the wall [1] (Newton’s
3rd Law). Since pressure = force/area, a pressure of the walls of the container is
experienced.
4