Question

A Carnot’s engine whose temperature of the source is 400K takes 200 calories
of heat at this temperature and rejects 150 calories of heat to the sink. (i)
What is the temperature of the sink. (ii) Calculate the efficiency of the engine

Answer 1

For Carnot engine
Q1/T1=Q2/T2
200/400=150/T
T=300K
And the efficiency of engine is
= [1-(lowertemp/higher temp)]*100
[1-(300/400)]*100= 25%

Answer 2

i)The temperature of sink is 300K

ii)Efficiency of engine is 25%

A Carnot engine is an ideal reversible heat engine operating between two temperatures T1(source) and T2(sink).

For a Carnot engine we have:--

Q2/Q1=T2/T1

Where Q1 is heat extracted from source

Q2 is heat rejected to the sink

T1 is temperature of the source

T2 is temperature of the sink.

Q1=200 calories Q2=150 calories T1=400K

i)so substituting value Q2/Q1=T2/T1

150/200=T2/400

T2=300K

ii) Efficiency of engine = (1-Q2/Q1)*100 or (1-T2/T1)*100

substituting values( 1-150/200)*100=1/4*100=25%