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Step-by-Step Solution
Step 1: Write down the given information
• A Carnot engine has an initial work-to-heat ratio W/Q = 1/6 (i.e., efficiency
$ \eta = \frac{W}{Q} = \frac{1}{6} $).
• When the sink (cold reservoir) temperature is lowered by 62°C, the new ratio becomes W/Q = 1/3
(i.e., efficiency $ \eta = \frac{1}{3} $).
• Let $T_1$ be the source (hot reservoir) temperature and $T_2$ be the sink (cold reservoir) temperature, both in Kelvin.
Step 2: Express the initial efficiency relation
For a Carnot engine, the efficiency is
$ \eta = 1 - \frac{T_2}{T_1} $.
Given $ \eta = \frac{1}{6} $, we have:
$ \frac{1}{6} = 1 - \frac{T_2}{T_1} $.
Rearranging,
$ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} $.
Therefore,
$ T_2 = \frac{5}{6} T_1 $.
Step 3: Express the new efficiency relation after the sink is lowered
After reducing sink temperature by 62°C, the new sink temperature becomes $T_2 - 62$ (in °C, but we will work in Kelvin correspondingly).
The new efficiency is given as
$ \eta = \frac{1}{3} $:
$ \frac{1}{3} = 1 - \frac{(T_2 - 62)}{T_1} $.
Step 4: Substitute $T_2 = \frac{5}{6} T_1$ into the new efficiency equation
From the initial calculation, $ T_2 = \frac{5}{6}T_1 $. Substitute this into the new efficiency relation:
$ \frac{1}{3} = 1 - \frac{\left(\frac{5}{6} T_1 - 62\right)}{T_1} $.
Simplifying:
$ \frac{1}{3} = 1 - \frac{5}{6} + \frac{62}{T_1} $.
So
$ \frac{1}{3} = \frac{1}{6} + \frac{62}{T_1} $.
Hence
$ \frac{1}{3} - \frac{1}{6} = \frac{62}{T_1} $.
$ \frac{1}{6} = \frac{62}{T_1} $.
Therefore
$ T_1 = 62 \times 6 = 372 \text{ K} $.
Step 5: Convert the source temperature to °C
Since 0 K corresponds to -273°C approximately:
$ T_1 (\text{in }^\circ\text{C}) = 372 - 273 = 99^\circ\text{C} $.
Step 6: Calculate the sink temperature
From
$ T_2 = \frac{5}{6} T_1 = \frac{5}{6} \times 372 = 310 \text{ K} $.
Converting to °C:
$ T_2 (\text{in }^\circ\text{C}) = 310 - 273 = 37^\circ\text{C} $.
Step 7: State the final results
Thus, the initial temperatures of the sink and source (in °C) are:
• Sink temperature = $37^\circ\text{C}$
• Source temperature = $99^\circ\text{C}$.
