Question

An ideal gas heat engine operate in Carnot cycle between \(327^{\circ }C\) and \(227^{\circ }C\). It absorbs \(6\times 10^{4}\) cal of heat at higher temperature. How much is the amount of heat converted in to work?

• A. \(2\times 10^{4}cal\)
• B. \(1\times 10^{4}cal\)
• C. \(4\times 10^{4}cal\)
• D. \(3\times 10^{4}cal\)

Hint:

Convert both temperatures to kelvin, find the Carnot efficiency, and multiply by the heat absorbed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Work done is \(W=\eta Q_{1}\) with \(\eta=1-\frac{T_{2}}{T_{1}}\).

Step 2: Temperatures
\(T_{1}=327+273=600\) K and \(T_{2}=227+273=500\) K.

Step 3: Calculation
\(\eta=1-\frac{500}{600}=\frac{1}{6}\)
\(W=\frac{1}{6}\times6\times10^{4}=1\times10^{4}\) cal

Step 4: Compare the options
This is option (B).

Final Answer:
The work obtained is 1 times 10 to the 4 cal. \[ \boxed{\text{(B) }1\times10^{4}\,\text{cal}} \]