Question

Carnot engine operating between temperatures \(T_1\) and \(T_2\) has efficiency \(0.2\). When \(T_2\) is lowered by \(45\) K, its efficiency becomes \(0.5\). Temperatures \(T_1\) and \(T_2\) are respectively

• A. \(150\) K, \(120\) K
• B. \(120\) K, \(150\) K
• C. \(60\) K, \(80\) K
• D. \(80\) K, \(60\) K

Hint:

Write efficiency = 1 - T2/T1 for both cases and solve.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The efficiency of a Carnot engine is \(\eta=1-\dfrac{T_2}{T_1}\).

Step 2: First condition:
\(0.2=1-\dfrac{T_2}{T_1}\), so \(T_2=0.8T_1\).

Step 3: Second condition:
The sink temperature is lowered to \(T_2-45\): \(0.5=1-\dfrac{T_2-45}{T_1}\), so \(T_2-45=0.5T_1\).

Step 4: Solve:
Substitute \(T_2=0.8T_1\): \(0.8T_1-45=0.5T_1\), so \(0.3T_1=45\) and \(T_1=150\) K. Then \(T_2=0.8\times150=120\) K. Option A.

Step 5: Why the other options are wrong.
B swaps source and sink (the source must be the hotter one). C and D give temperatures that do not satisfy \(\eta=0.2\) either.

Final Answer:
T1 = 150 K and T2 = 120 K. \[ \boxed{\text{(A) }150\ \text{K},\ 120\ \text{K}} \]