Question

Let \(\eta _{1}\) be the efficiency of a Carnot engine at \(T_{H}=547^{\circ }\text{C}\) and \(T_{C}=247^{\circ }\text{C}\) while \(\eta _{2}\) is the efficiency at \(T_{H}=847^{\circ }\text{C}\) and \(T_{C}=147^{\circ }\text{C}\). If \(\eta _{1}=\frac{x}{41}\eta _{2}\), the value of \(x\) is

• A. \(24\)
• B. \(25\)
• C. \(20\)
• D. \(22\)

Hint:

Convert the temperatures to kelvin, compute each efficiency as 1 minus Tc over Th, and compare.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The efficiency of a Carnot engine is \(\eta=1-\dfrac{T_{C}}{T_{H}}\), with temperatures in kelvin.

Step 2: First engine
\(T_{H}=820\ \text{K}\), \(T_{C}=520\ \text{K}\). \(\eta_{1}=1-\dfrac{520}{820}=\dfrac{300}{820}=\dfrac{15}{41}\).

Step 3: Second engine
\(T_{H}=1120\ \text{K}\), \(T_{C}=420\ \text{K}\). \(\eta_{2}=1-\dfrac{420}{1120}=\dfrac{700}{1120}=\dfrac{5}{8}\).

Step 4: Compare
\(\dfrac{\eta_{1}}{\eta_{2}}=\dfrac{15}{41}\times\dfrac{8}{5}=\dfrac{24}{41}\), so \(\eta_{1}=\dfrac{24}{41}\eta_{2}\) and \(x=24\).

Final Answer:
The value of x is 24. \[ \boxed{\text{(A) }24} \]