Problem 11.11. The low temperature reservoir of a Carnot engine is at 7°C and has an efficiency of 40%. It is desired to increase the efficiency to 50%. By how much degrees the temperature of hot reservoir be increased.
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The low temperature reservoir of a Carnot engine is at 7°C and has an efficiency of 40%. It is desired to increase the efficiency to 50%. By how much degrees the temperature of hot reservoir be increased.?
Data
T2=7°c+273=280k
E1=40%
E2=50%
Increase in temp of hot body= ΔT=?
Solution
E1=(1-t2/t1)×100
40=(1-280/t1)×100
40/100=t1 - 280/t1
2/5/= t1 -280/= 2t1
5(t1 -280) =2t1
5t1 -2t1= 1400
3t1=1400
T1 =466.6k. Ans 1
E2=(1 -t2/t1) × 100
50=(1 -280)/t1 ×100
50/100b=t1 +280/t1
1/2=t¹-280) t¹
2t¹× 280= t1
560k
Now increase in temperature of hot body is
ΔT =t1 -t1
ΔT =560 -466.6k
ΔT =93.4 k......
Answer
