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?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_c}{T_h}$, where $T_c$ and $T_h$ are the temperatures of the cold and hot reservoirs, respectively, in Kelvin. To convert the given temperatures to Kelvin, add 273.15 to each: $T_h = 527 + 273.15 = 800.15 K$ and $T_c = -73 + 273.15 = 200.15 K$. Thus, $\eta = 1 - \frac{200.15}{800.15}$. The work done (W) is related to the heat extracted from the hot reservoir (Q_h) by the efficiency: $W = \eta \cdot Q_h$. Rearranging for $Q_h$, we get $Q_h = \frac{W}{\eta}$. Substituting the values, we find $Q_h = \frac{1000}{1 - \frac{200.15}{800.15}}$, which calculates to approximately 1333 J.

A Carnot Engine Operates Between Hot and Cold Reservoirs with Temperatures