Question

An engine operating between temperatures \(T_1\) and \(T_2\) has efficiency \(\frac{1}{5}\). When \(T_2\) is lowered by 45 K , its efficiency becomes \(\frac{1}{2}\). Temperatures \(T_1\) and \(T_2\) are respectively

• A. \(100\text{ }\mathrm{K},70\text{ }\mathrm{K}\)
• B. \(160\text{ }\mathrm{K},120\text{ }\mathrm{K}\)
• C. \(140\text{ }\mathrm{K},110\text{ }\mathrm{K}\)
• D. \(150\text{ }\mathrm{K},120\text{ }\mathrm{K}\)

Hint:

Use the efficiency of an ideal engine in terms of the two temperatures, written for both the initial and the changed sink temperature.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Efficiency of an engine between \(T_1\) (source) and \(T_2\) (sink) is \(\eta=1-\dfrac{T_2}{T_1}\).

Step 2: First condition
\(\dfrac15=1-\dfrac{T_2}{T_1}\) gives \(T_2=\dfrac45T_1\).

Step 3: Second condition
\(\dfrac12=1-\dfrac{T_2-45}{T_1}\) gives \(T_2-45=\dfrac12T_1\).

Step 4: Solve
Substituting: \(\dfrac45T_1-45=\dfrac12T_1\), so \(\dfrac{3}{10}T_1=45\) and \(T_1=150\text{ K}\).
Then \(T_2=\dfrac45\times150=120\text{ K}\).

Final Answer:
\(T_1=150\text{ K}\) and \(T_2=120\text{ K}\). \[ \boxed{\text{(D) }\text{150 K, 120 K}} \]