A Level Specific Heat Capacity Questions
Q1. A domestic room heater which uses natural gas as the fuel, produces an output power of 4.5 kW.
The energy obtained from a cubic metre of this fuel is 39 MJ. The density of the gas is 0.72 kg m–3 at atmospheric pressure.
(a) Show that the volume of gas which must be burnt to produce a steady output is 6.9 x 10–3
m3
each minute.
energy produced per minute = 4.5 x 103 x 60 
= 2.7 x 105 J
volume of gas required = (2.7 x 105 )/(39 x 106) m3) = 6.9 10–3 m3
OR
volume of gas per second = 1.15 x 10–4m3
volume of gas per minute = 1.15 x 10–4 x 60 m3 = 6.9 x 10–3 m3
(2 marks)
(b) Calculate
(i) the mass of gas which is burnt each minute,
= m/V so m = V
m = 0.72 x 6.9 x 10–3
= 5.0 x 10–3 kg
(ii) the number of molecules of natural gas which pass through the burner each minute if the molar mass of the gas is 1.6 x 10–2 kg mo1–1.
mass of one molecule = 1.6 x 10-2/NA
= 2.66 x 10–26 kg
number of molecules =(5.0 x 10-3)/(2.66 x 10–26)
= 1.9 x 1023
OR
number of moles = (5.0 x 10-3)/(1.6 x 10-2) = 0.31
number of molecules (= 0.31 x 6.0 x 1023) = 1.9 x 10
23
(4 marks)
(c) When the heater is first turned on, the temperature of the air entering it is 14°C and the temperature of the air leaving it is 36°C. If the specific heat capacity of air is 990 J kg–1 K–1, calculate the mass of air passing through the heater in one minute.
energy per minute (from (a)) = 2.7 x 105 J
Q = mc 
m = Q/(c )
= 2.7 x 105/(990 x (36 - 14))
= 12(.4) kg
OR
m = 4500/(990 x (36 - 14)) for 1 second
= 0.21 (kg for 1 second)
For 1 minute m= 0.21 x 60) = 12 kg
(3 marks)
(d) The room could have been heated using a 4.5 kW electric heater. Explain with a suitable calculation why such a heater could not be operated from a 13 A, 230 V socket.
P = IV
so, I = P/V
=4.5 x 103/230
= 19(.6) A
This means that the current required for 4.5 kW would exceed fuse rating and blow the fuse when switched on.
(2 marks)
(Total 11 marks)
About this site
