First Thermodynamics Examples Sheet .. and more
1. If the specific heat of water is 4.2 kJ kg-1 C-1 and the acceleration of gravity is
9.81 m s-2, estimate the temperature difference between the top and bottom of the
Pistyll Rheadr waterfall in Wales (height 73 m). (Hint: consider 1 kg of water
negotiating the fall).
Discuss any factors which might affect the accuracy of this answer.
2. Assuming that 80% of the energy supplied to a 1.5-kw electric kettle goes to heat
the water in it, how long will the kettle take to heat one litre of water from 20 C to 100
C?
3. It is found that a 60-horsepower laboratory petrol engine will only run for 15
minutes at full power on its full tank of 2 gallons of petrol whose density is 0.8 kg/litre
and calorific value 42 000 kJ/kg. You are also given that 1 gallon = 4.7 litres and that
1 horsepower = 0.746 kw.
Calculate
(i)
The thermal energy released by burning the 2 gallons of
petrol.
(ii)
The mechanical energy delivered in the same time.
(iii)
The efficiency of the engine.
3. Determine the quantities Q, W and E for the following operations. You are given
the following:
PV = mRT, where R = 287 J kg-1 C-1 kg-1 for air
Cp = 1.007 kJ kg-1 C-1
Cv = 0.72 kJ kg-1 C-1
Remember - ADIABATIC means no heat is added or removed
CONSTANT VOLUME means no work is done
ISOTHERMAL means no change in internal energy E
(a)
1 kg of air is heated from 20 C to 45 C at constant volume.
(b)
As (a) but at constant pressure.
[(a) Q = E = 18 kJ, W = 0; (b) Q = 25.175 kJ, E = 18 kJ, W = 7.175 kJ)]
4. 1 kg of air expands isothermally at 18 C from 100 000 Pa to 50 000 Pa. Determine
its final volume, using PV = mRT (remember T is in degree ABSOLUTE)
[Some useful relations (NB - all for non-flow processes)
These relations all apply per kilogramme of the "working fluid"
Constant volume
Work
E
Q
Zero
Cv(T2 - T1)
Cv(T2 - T1)
Cv(T2 - T1)
Cp(T2 - T1)
Constant pressure R(T2 - T1)
For constant-pressure, the work can also be calculated as the pressure times the
change in volume. It can be calculated as pdv for all processes (which is how the
polytropic expressions were calculated) but that is rather hard work!
Also: PV = mRT (General Gas Equation - R = 287 J kg-1 K-1
The following two apply to any constant mass of gas:

Adiabatic changes: PV = constant
Isothermal changes: PV = constant
An example of their use
A single-cylinder engine operates on the following series of events (or cycle).
1.
2.
3.
4.
5.
250 ml of air at 20 C and atmospheric pressure (98 100 Pa) are ingested.
This air is compressed adiabatically to a volume of 30 ml.
It is now heated at constant volume by the addition of 0.058 kJ from the
combustion of the fuel.
The air is now allowed to expand adiabatically back to a volume of 30 ml.
The air and spent fuel are now exhausted to atmosphere, undoubtedly
polluting it..
For the compression, we need to know the mass of the air and the pressure and
temperature reached. The mass is easily calculable by PV = mRT ..
giving m = PV/RT.
98100  0.25  10-3 = m  287  293, so m = .292  10-3 kg


We can also easily calculate the final pressure from P1V1 = P2V2 and P2 =

P1(V1/V2) . It started at 250 ml and 98 100 Pa and finished at 30 ml, so...
P2 = 98 100(250/30)1.4 pascal = 1.91 MPa
and the temperature from PV = mRT:
T2 = PV/(mR) = 1.91  106  30 10-6 / (.292  10-3  287) = 684 K
so the internal energy change will be
mCv(T2 - T1) = 0.72  .292  10-3  (684 - 293) = 0.082 kJ
which equals the work done on the system, as no heat exchange occurred.
This next section in italics calculates the work done from first principles to convince
me that the answer is right.
The work done is calculated by pdv given that pvn = constant ... n = 1.4 for air
and we can work out the constant from 98100  (0.25  10-3)1.4 = 0.889.
So p=0.889/v1.4 = 0.889v-1.4.
By substitution, the integral becomes pdv = (0.889v-1.4)dv = [-0.889/0.4  v-0.4]
between the two volumes 250 ml (250  10-6 m3) and 30 ml (30  10-6 m3). If we
substitute these values in the usual way for definite integrals, we obtain:
-0.889/0.4  [(30  10-6)-0.4 - (250  10-6)-0.4] = 82 J = 0.082 kJ (actually minus
because the work was done on the system, not by it).
As the compression was adiabatic, Q = 0, so U must also be 0.082 kJ. This gives a
final temperature of T2, where:
U = Cv(T2 - T1) = 0.72  .292  10-3  (T2 - 293) = .082
giving T2 = 682 K.}
The energy is now added at constant volume by the fuel burning.
Q = 0.058 kJ, so the temperature will rise by 0.058/(0.72 .292  10-3) = 276 K,
giving a final temperature of 276 + 682 = 956 K.
The heat from the fuel was added at constant volume, so the resulting pressure can
be calculated from PV = mRT.
P = .292  10-3  287  956/(30  10-6) = 2.67 MPa
We can now calculate the final pressure and temperature at the end of the adiabatic
expansion stage of the cycle (the 'power stroke').
The 'before and after' volumes are now 30 ml and 250 ml, so the final pressure will
be:

P2 = P1(V1/V2) = 2.67  106  (30/250)1.4 = 137 200 Pa
and we calculate the temperature by PV = mRT:
T
=
PV/(mR)
=
=
137 200  250  10-6/(.292  10-3  287)
409 K
So the loss of internal energy is mCv(T1 - T2)
= .292  10-3  0.72  (956-409)
= 0.115 kJ, which is the work done in the expansion.
We can calculate the efficiency from these figures.
Heat supplied = 0.058 kJ
Work done = (0.114 - .082) = 0.032 kJ (note that we had to subtract the work done
on the gas during the compression!)
So it looks as if the efficiency is 0.032/0.058 = 55 % !
In fact it will not be as high as this, because the engineering limitations will prevent
us from using such a large heat input.
We (you?) will calculate what the answer would have been if the heat from the fuel
were only 0.03 kJ.