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Step-by-Step Solution
Step 1: Convert the temperatures to Kelvin
The Carnot engine operates between 227°C and 127°C. To use these in thermodynamic
formulas, convert them into the Kelvin scale by adding 273.
$T_1 = 227 + 273 = 500\text{ K}$
$T_2 = 127 + 273 = 400\text{ K}$
Step 2: Recall the efficiency relation for a Carnot engine
The efficiency $ \eta $ of a Carnot engine is given by:
$ \eta = 1 - \frac{T_2}{T_1} $
Also, this efficiency can be expressed as:
$ \eta = 1 - \frac{Q_2}{Q_1} $
Here, $Q_1$ is the heat absorbed at the higher temperature $T_1,$ and $Q_2$ is the heat released at the lower temperature $T_2$.
Step 3: Equate the two expressions for efficiency
$ 1 - \frac{T_2}{T_1} = 1 - \frac{Q_2}{Q_1} $
Substitute $T_1 = 500\text{ K}, T_2 = 400\text{ K}, \text{ and } Q_1 = 6 \times 10^4 \text{ cal}:$
$ 1 - \frac{400}{500} = 1 - \frac{Q_2}{6 \times 10^4} $
Step 4: Solve for $Q_2$
$ 1 - \frac{400}{500} = 1 - \frac{Q_2}{6 \times 10^4} $
$ 1 - 0.8 = 1 - \frac{Q_2}{6 \times 10^4} $
$ 0.2 = 1 - \frac{Q_2}{6 \times 10^4} \quad \Longrightarrow \quad \frac{Q_2}{6 \times 10^4} = 0.8 \quad \Longrightarrow \quad Q_2 = 4.8 \times 10^4 \text{ cal} $
Step 5: Calculate the work done
The net heat converted to work, $W,$ is the difference between the heat absorbed ($Q_1$) and the heat released ($Q_2$):
$ W = Q_1 - Q_2 = 6 \times 10^4 \text{ cal} - 4.8 \times 10^4 \text{ cal} $
$ W = 1.2 \times 10^4 \text{ cal} $
Thus, the amount of heat converted into work by this Carnot engine is $1.2 \times 10^4$ cal.
