An ideal Carnot engine extracts 100 J from a heat source and dumps 40 J to a heat sink at 300 K. The temperature of the heat source is
750 K
An ideal Carnot engine operates based on the Carnot cycle, which is the most efficient thermodynamic cycle possible. A key property of a Carnot engine is the relationship between the heat exchanged with the hot and cold reservoirs and their absolute temperatures.
The problem provides the following information for an ideal Carnot engine:
We need to find the temperature of the heat source ($T_H$).
For any ideal Carnot engine, the ratio of the heat rejected to the cold reservoir ($Q_C$) to the heat absorbed from the hot reservoir ($Q_H$) is equal to the ratio of the absolute temperatures of the cold reservoir ($T_C$) and the hot reservoir ($T_H$). This relationship is given by the formula:
$$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$
We can substitute the given values into the formula to find the unknown heat source temperature, $T_H$:
$$\frac{40 \, \text{J}}{100 \, \text{J}} = \frac{300 \, \text{K}}{T_H}$$
Simplify the fraction on the left side:
$$\frac{4}{10} = \frac{300 \, \text{K}}{T_H}$$
$$\frac{2}{5} = \frac{300 \, \text{K}}{T_H}$$
Now, we can solve for $T_H$ by cross-multiplication or by rearranging the equation:
$$T_H = \frac{300 \, \text{K} \times 5}{2}$$
$$T_H = \frac{1500 \, \text{K}}{2}$$
$$T_H = 750 \, \text{K}$$
Thus, the temperature of the heat source is 750 K.
This calculation shows how the performance of an ideal Carnot engine, in terms of heat exchange, is directly related to the temperatures of the heat source and heat sink.
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