Question

The efficiency of a Carnot cycle is \( \frac{1}{6} \). By lowering the temperature of the sink by 65 K, it increases to \( \frac{1}{3} \). The initial and final temperature of the sink are:

• A. \( 400 \text{ K}, 310 \text{ K} \)
• B. \( 525 \text{ K}, 65 \text{ K} \)
• C. \( 309 \text{ K}, 235 \text{ K} \)
• D. \( 325 \text{ K}, 260 \text{ K} \)

Hint:

For Carnot cycle problems, always use the efficiency formula \( \eta = 1 - \frac{T_C}{T_H} \) and systematically solve for the unknowns by forming and solving equations.

Answer 1

The Correct Option is D

To solve the problem, we need to understand the concept of Carnot cycle efficiency. The efficiency (\( \eta \)) of a Carnot engine is given by the formula:

\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
The problem provides the efficiency as \( \frac{1}{6} \), which implies:

\[\frac{1}{6} = 1 - \frac{T_{\text{sink, initial}}}{T_{\text{source}}}\]
Simplifying, we find:
\[\frac{T_{\text{sink, initial}}}{T_{\text{source}}} = 1 - \frac{1}{6} = \frac{5}{6}\]
Thus, \( T_{\text{source}} = \frac{6}{5} \times T_{\text{sink, initial}} \).

When the efficiency increases to \( \frac{1}{3} \) due to the sink temperature dropping by 65 K:

\[\frac{1}{3} = 1 - \frac{T_{\text{sink, final}}}{T_{\text{source}}}\]
Let's simplify:
\[\frac{T_{\text{sink, final}}}{T_{\text{source}}} = 1 - \frac{1}{3} = \frac{2}{3}\]
Therefore:
\( T_{\text{source}} = \frac{3}{2} \times T_{\text{sink, final}} \)

We now equate the expressions for \( T_{\text{source}} \):
\[\frac{6}{5} \times T_{\text{sink, initial}} = \frac{3}{2} \times T_{\text{sink, final}}\]
\[\frac{6}{5} T_{\text{sink, initial}} = \frac{3}{2} (T_{\text{sink, initial}} - 65)\]
Solve for \( T_{\text{sink, initial}} \):
\[6 T_{\text{sink, initial}} = 7.5(T_{\text{sink, initial}} - 65)\]
\[6 T_{\text{sink, initial}} = 7.5 T_{\text{sink, initial}} - 487.5\]
\[1.5 T_{\text{sink, initial}} = 487.5\]
\[T_{\text{sink, initial}} = \frac{487.5}{1.5} = 325 \text{ K}\]
\[T_{\text{sink, final}} = 325 \text{ K} - 65 \text{ K} = 260 \text{ K}\]

Thus, the initial and final temperatures of the sink are \( 325 \text{ K} \) and \( 260 \text{ K} \), respectively.

Answer 2

Step 1: Understanding Carnot Efficiency Formula The efficiency \( \eta \) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_C \) is the sink (low-temperature reservoir) temperature, and \( T_H \) is the source (high-temperature reservoir) temperature. 
Step 2: Setting Up Equations for Given Conditions We are given that initially: \[ \frac{1}{6} = 1 - \frac{T_C}{T_H} \] Rearranging, \[ \frac{T_C}{T_H} = 1 - \frac{1}{6} = \frac{5}{6} \] Thus, \[ T_C = \frac{5}{6} T_H \] After lowering the sink temperature by 65 K, the efficiency increases to \( \frac{1}{3} \), so: \[ \frac{1}{3} = 1 - \frac{T_C - 65}{T_H} \] \[ \frac{T_C - 65}{T_H} = 1 - \frac{1}{3} = \frac{2}{3} \] Thus, \[ T_C - 65 = \frac{2}{3} T_H \] 
Step 3: Solving for \( T_H \) and \( T_C \) From the two equations: 1. \( T_C = \frac{5}{6} T_H \) 2. \( T_C - 65 = \frac{2}{3} T_H \) Substituting the first equation into the second: \[ \frac{5}{6} T_H - 65 = \frac{2}{3} T_H \] Multiplying everything by 6 to clear fractions: \[ 5 T_H - 390 = 4 T_H \] \[ T_H = 390 \text{ K} \] Now, substituting into \( T_C = \frac{5}{6} T_H \): \[ T_C = \frac{5}{6} \times 390 = 325 \text{ K} \] After decreasing by 65 K: \[ T_C' = 325 - 65 = 260 \text{ K} \] Thus, the initial and final sink temperatures are 325 K and 260 K, respectively.