Question

in a heat engine,the temp. of source and sink are 500K and 375K . if the engine consumes 25 * 10^5 J/sec. find
(I) efficiency of the engine.
(ii) work done per cycle.
(iii) heat rejected to the sink.

Answer 1

Answer:

(I) 0.25

(ii)25×10^5J

(iii)75×10^5J

Explanation:

(I)efficiency=1-(T2/T1)

where, T2=temperature of sink

T1=temperature of source

Efficiency=1-(375/500)=0.25

(ii) work done=energy consumed by

engine=25×10^5J

(iii)Q1=W+Q2

where, Q1=heat released from source

Q2=heat rejected to sink

W=work done by engine

Efficiency=W/Q1=0.25

Q1=W×4=100×10^5

Q2=Q1-W=75×10^5