A Carnot engine operates between hot and cold reservoirs with temperatures `527^@ C and -73.0^@ C`, respectively. If the engine performs 1000.0 J of work per cycle, how much heat is extracted per cycle from the hot reservoir?

Options

A. 878J
B. 133J
C. 1163 J
D. 1527 J

Answer

To solve the problem of how much heat is extracted per cycle from the hot reservoir by a Carnot engine operating between two temperatures, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures of the hot and cold reservoirs are given in degrees Celsius. We need to convert these temperatures to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For the hot reservoir:
\[ T_H = 527°C + 273.15 = 800.15 K \]
- For the cold reservoir:
\[ T_C = -73°C + 273.15 = 200.15 K \]

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]

Substituting the values we found:
\[ \eta = 1 - \frac{200.15}{800.15} \]

Calculating this gives:
\[ \eta \approx 1 - 0.250 = 0.75 \]
So, the efficiency is 75% or \( \frac{3}{4} \).

### Step 3: Use the efficiency to find the heat extracted from the hot reservoir
The efficiency of a Carnot engine can also be expressed in terms of work done (W) and heat extracted from the hot reservoir (Q_H):
\[ \eta = \frac{W}{Q_H} \]

Rearranging this gives:
\[ Q_H = \frac{W}{\eta} \]

Given that the work done (W) is 1000 J, we can substitute the values:
\[ Q_H = \frac{1000 J}{0.75} \]

Calculating this gives:
\[ Q_H = \frac{1000}{0.75} = 1333.33 J \]

### Conclusion
The heat extracted per cycle from the hot reservoir is approximately **1333.33 J**.

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