A Carnot engine operate between 800K and 200K.if it absorbs 8kJ of heat in each cycle,then the work done by it per cycle is
1. 1kJ
2. 2kJ
3. 2.7 kJ
4. 6 kJ
Answers
Answered by
14
efficiency= T2-T1/T2=work output/work input
here
T1: source temperature
T2: sink temperature
n(eta)=800-200/800 =0.75
now, n=energy used up/energy provided
and remember that energy used up=work done..
so, n=work output/work input
0.75=8kJ-xkJ/8kJ
(xkJ=work done)
0.75×8=8-x
6-8=-x
therefore, x=2
hence work done=2kJ
here
T1: source temperature
T2: sink temperature
n(eta)=800-200/800 =0.75
now, n=energy used up/energy provided
and remember that energy used up=work done..
so, n=work output/work input
0.75=8kJ-xkJ/8kJ
(xkJ=work done)
0.75×8=8-x
6-8=-x
therefore, x=2
hence work done=2kJ
Answered by
3
2 one is your answer
Similar questions