Lesson Plan

Explain how a heat engine converts heat into work using a cyclical process, and derive W = Q_h − Q_c from the first law

Explain why the second law of thermodynamics limits every heat engine to less than 100% efficiency

Explain how heat pumps, refrigerators, and air conditioners move heat from cold to hot using work input (Q_h = Q_c + W)

Calculate thermal efficiency, Eff = W/Q_h, and solve problems for work output and efficiency

12.4

"One of the most important things we can do with heat is to use it to do work for us. A heat engine does exactly this—it makes use of the properties of thermodynamics to transform heat into work." "A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. All heat engines use cyclical processes." "Because there is no change in internal energy for a complete cycle ($\Delta U = 0$), we have $0 = Q - W$, so that $W = Q.$ … $W = Q_h - Q_c$ for a cyclical process." "According to the second law of thermodynamics, heat engines cannot have perfect conversion of heat into work. … there is a minimum amount of $Q_h$ that cannot be used for work." "Heat pumps, air conditioners, and refrigerators utilize heat transfer of energy from low to high temperatures, which is the opposite of what heat engines do. … the total heat transfer to the hot reservoir is $Q_h = Q_c + W.$" "We define thermal efficiency, Eff, to be the ratio of useful energy output to the energy input … $Eff = \frac{W}{Q_h}.$ An efficiency of 1, or 100 percent, would be possible only if there were no heat to the environment ($Q_c = 0$)." "All values of heat ($Q_h$ and $Q_c$) are positive; there is no such thing as negative heat. The direction of heat is indicated by a plus or minus sign." "(a) What is the work done by the power station? … $W = 2.50 \times 10^{14}\text{ J} - 1.48 \times 10^{14}\text{ J} = 1.02 \times 10^{14}\text{ J}.$ (b) … $Eff = \frac{1.02 \times 10^{14}\text{ J}}{2.50 \times 10^{14}\text{ J}} = 0.408,\text{ or } 40.8%.$"