Question

A reversible engine with an efficiency of \(30.0 \%\) has $T_{\mathrm{C}}=310.0 \mathrm{K} .\( (a) What is \)T_{\mathrm{H}} ?$ (b) How much heat is exhausted for every \(0.100 \mathrm{kJ}\) of work done?

Short answer

Answer: The hot temperature reservoir is approximately 442.9 K, and the amount of heat exhausted for every 0.100 kJ of work done is approximately 0.233 kJ.

Step-by-step solution

Use the formula for efficiency of a reversible engine

The efficiency of a reversible engine, \(\eta\), is related to the hot and cold temperature reservoirs (\(T_H\) and \(T_C\)) by the equation: $$ \eta = 1 - \frac{T_C}{T_H} $$ We are given the efficiency, \(\eta\), as \(30.0\%\) or \(0.30\), and we're given the cold temperature reservoir, \(T_C = 310.0 K\). We need to solve for the hot temperature reservoir, \(T_H\). Rearrange the equation above to solve for \(T_H\): $$ T_H = \frac{T_C}{1 - \eta} $$

Calculate \(T_H\) using the efficiency and \(T_C\) values

Now, plug in the given values of efficiency, \(\eta = 0.30\), and the cold temperature reservoir, \(T_C = 310.0 K\), into the above equation: $$ T_H = \frac{310.0}{1 - 0.30} = \frac{310.0}{0.70} $$ Upon evaluating the expression, we get the hot temperature reservoir, \(T_H \approx 442.9 K\). So, \(T_H = 442.9 K\).

Find the heat exhausted for every \(0.100 kJ\) of work done

To find the heat exhausted for every \(0.100 kJ\) of work done, first, we need to find how much heat is supplied to the engine, represented by \(Q_H\). We use the formula, $$ W = \eta Q_H $$ Rearrange the equation above to solve for \(Q_H\): $$ Q_H = \frac{W}{\eta} $$ Given the work done, \(W = 0.100 kJ\) and efficiency, \(\eta = 0.30\). $$ Q_H = \frac{0.100}{0.30} = 0.333... kJ $$

Find the heat exhausted, \(Q_C\), using the energy conservation principle

Now, we'll use the energy conservation principle, which states that the total heat supplied to the engine, \(Q_H\), is equal to the sum of the heat exhausted, \(Q_C\), and the work done, \(W\). Rearrange the equation as: $$ Q_C = Q_H - W $$ Substitute the value of \(Q_H = 0.333... kJ\) and \(W = 0.100 kJ\) into the equation: $$ Q_C = 0.333... - 0.100 = 0.233... kJ $$ So, the heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\). In summary, the hot temperature reservoir, \(T_H\), is approximately \(442.9 K\) and the amount of heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\).