Question

A heat engine operating between ${227}^0$C and ${27}^0$C absorbs $2$ kcal of heat from the ${227}^0$C reservoir as per cycle. The amount of work done in one cycle is:

• A. $0.4$ kcal
• B. $0.8$ kcal
• C. $4$ kcal
• D. $8$ kcal

Answer 1

Answer: (B)

$0.8$ kcal

$T_1=227℃=(227+273)K=500K$

$T_2=27℃=(27+273)=300K$

$\text{Heat absorbed}\left(q\right)=2kcal$

$\text{work output}\left(w\right)=?$

As we know that,

$Efficiency\left(\eta \right)=1-\frac{T_2}{T_1}$

$\Rightarrow \eta =1-\frac{300}{500}=\frac{200}{500}$

$\eta =0.4⟶\left(1\right)$

Again,

$\eta =\frac{w}{q}$

$\Rightarrow \eta =\frac{w}{2}⟶\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{w}{2}=0.4$

$\Rightarrow w=0.8kcal$

Hence, amount of work done in 1 cycle = 0.8 kcal.