A thermodynamic engine with an efficiency of 30% performs 2500 J of work.How much thermal energy is discharged to the lower temperature reservoir?

Question

Quizplus · Accepted Answer

The efficiency ($\eta$) of a thermodynamic engine is given by the ratio of the work output ($W$) to the heat input ($Q_{in}$). Mathematically, $\eta = \frac{W}{Q_{in}}$. Given $\eta = 30\% = 0.30$ and $W = 2500 J$, we can solve for $Q_{in}$ as follows: $0.30 = \frac{2500}{Q_{in}}$, which gives $Q_{in} = \frac{2500}{0.30} = 8333.33 J$. The thermal energy discharged ($Q_{out}$) is the difference between the heat input and the work output, $Q_{out} = Q_{in} - W = 8333.33 J - 2500 J = 5833.33 J$, which is approximately 5800 J.

A Thermodynamic Engine with an Efficiency of 30% Performs 2500