Search any question & find its solution
Question:
Answered & Verified by Expert
A Carnot engine absorbs $6 \times 10^5 \mathrm{cal}$ at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is
Options:
Solution:
1472 Upvotes
Verified Answer
The correct answer is:
$5 \times 10^5 \mathrm{~J}$
Here, $\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}$
$\begin{aligned} & \mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K} \\ & \mathrm{~T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\end{aligned}$
Work done/cycle, $\mathrm{W}=$ ?
As $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ or
$\mathrm{Q}_2=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \times \mathrm{Q}_1=\frac{400}{500} \times 6 \times 10^5$
$=4.8 \times 10^5 \mathrm{cal}$
$\begin{aligned} \text { As, } & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \\ \mathrm{~W} & =1.2 \times 10^5 \mathrm{cal}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \\ \mathrm{~W} & =5.04 \times 10^5 \mathrm{~J}\end{aligned}$
$\begin{aligned} & \mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K} \\ & \mathrm{~T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\end{aligned}$
Work done/cycle, $\mathrm{W}=$ ?
As $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ or
$\mathrm{Q}_2=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \times \mathrm{Q}_1=\frac{400}{500} \times 6 \times 10^5$
$=4.8 \times 10^5 \mathrm{cal}$
$\begin{aligned} \text { As, } & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \\ \mathrm{~W} & =1.2 \times 10^5 \mathrm{cal}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \\ \mathrm{~W} & =5.04 \times 10^5 \mathrm{~J}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.