Question

A heat engine operating between 227 deg C and 27 deg C absorbs 1 kcal of heat from the 227 deg C reservoir per cycle. Calculate
(1) the amount of heat discharged into the low temperature reservoir.
(2) the amount of work done per cycle.
(3) the efficiency of cycle.

• A. 0.4 kcal, 0.6 kcal, 40%
• B. 0.6 kcal, 0.4 kcal, 40%
• C. 0.4 kcal, 0.6 kcal, 60%
• D. 0.7 kcal, 0.4 kcal, 40%

Answer 1

Answer: (A)

0.4 kcal, 0.6 kcal, 40%

Calculation of work done:
$T^{\prime }=227+273=500$ K
$T=27+273=300$ K
$q^{\prime }=1$ 1 kcal
$\frac{w}{q^{\prime }}=\frac{T^{\prime }-T}{T^{\prime }}$
$w=q\left(\frac{T^{\prime }-T}{T^{\prime }}\right)$
Putting the various values in the above reaction
$w=1\left(\frac{500-300}{500}\right)=1\times \frac{200}{500}=0.4$ kcal.
Calculation of heat discharged q
$q^{\prime }-q=w$
$1-q=0.4$
$q=0.6$ kcal
Calculation of efficiency
$\eta =\frac{w}{q^{\prime }}=0.4$ kcal or 40%