Efficiency, redefined

A heat engine takes heat $Q_H$ from the hot reservoir, does useful work $W$ and releases heat $Q_C$ to the cold reservoir. Hence the efficiency of the heat engine is $\frac{W}{Q_H} = \frac{Q_H - Q_L}{Q_H}$

For the maximal efficiency (a Carnot cycle), this becomes $\frac{T_H - T_L}{T_H} = \frac{343.15-283.15}{343.15} = 0.17485$

A heat pump, on the other hand, does work $W$ to pump heat $Q_C$ into the hot reservoir $Q_H$ (this can be considered the 'useful heat')

Hence the efficiency of the heat pump is $\frac{Q_H}{W} = \frac{Q_H}{Q_H - Q_L}$

For the maximal efficiency (a Carnot cycle), this becomes $\frac{T_H}{T_H - T_L} = \frac{343.15}{343.15-283.15} = 5.7192$

The difference is $5.7192 - 0.17485 = 5.54$

k, the coefficient of performance, is defined to be Q {h}/W. To find k in terms of T {c} and T {h}, we realize that W = Q {h} - Q {c} from the first law of thermodynamics. By substitution, we get \frac{Q {h}}{Q {h}-Q {c}}. Assuming Cornot efficiency, we have \frac{Q {h}}{Q {c}} = \frac{T {h}}{T {c}} where the temperatures are absolute temperatures (that is they are in Kelvin). From substitution can obtain \frac{T {h}}{T {h}-T_{c}} for the maximum theoretical efficiency for the heat pump.

For the heat engine, the efficiency (e) is defined to be \frac{W {out}}{W {in}}. The Work in is equal to Q {h} since for a heat engine we add heat to make it do more work. Also, we notice that for a heat engine, by energy conservation, that W {out} = Q {h}-Q {c}. By substitution we obtain e = \frac{Q {h}-Q {c}}{Q {h}} = 1 - \fract{Q {c}}{Q {h}} and assuming Carnot efficiency we obtain e = 1 - \frac{T {c}}{T_{h}}.

Thus k - e_{heat engine} = \frac{343.15 K}{343.15 - 283.15 K} - (1 - \frac{283.15 K}{343.15 K}) = 5.54

Given T(h)=70deg=(70+273.15)K=343.15K T(c)=10deg=(10+273.15)K=283.15K The maximal efficiency of heat engine is given by 1-T(c)/T(h). The maximal efficiency(COP) of heat pump is given by T(h)/[T(h)-T(c)] since heat is delivered to a hot reservoir at temperature T(h).

Plugging the values of T(h) and T(c) into both eq yield the difference as 5.719-0.175=5.54

The maximal efficiency of the heat pump can be written as $[COP]=\frac {T_{h}}{T_{h}-T_{c}}$ .
The maximal efficiency of the heat engine can be written as $\eta=\frac {T_{h}-T_{c}}{T_{h}}$ .
So, the difference in the maximal effieciency of the heat pump and the heat engine is $D=\frac {T_{h}}{T_{h}-T_{c}}-\frac {T_{h}-T_{c}}{T_{h}}$ .
Solving we get $D=5.544$ .

Since we are asked for the difference in maximal efficiency, we should use the efficiency of a Carnot cycle which theoretically gives the highest efficiency.

Heat engine takes in the heat from the hot reservoir and delivers work. Therefore, its efficiency is $\eta_1 = \frac{W}{Q_h}$ . Since $W = Q_h-Q_c$ due to the conservation of energy, we have that $\eta_1 = \frac{Q_h-Q_c}{Q_h}$ which is for a Carnot cycle $\eta_1 = \frac{T_h-T_c}{T_h} = 0.174$ .

Heat pump takes in work and gives out heat. Therefore, $\eta_2 = \frac{Q_h}{W} = \frac{Q_h}{Q_h - Q_c} = \frac{T_h}{T_h-T_c} = 5.72$ .

The difference in maximal theoretical efficiency is $\eta_2 - \eta_1 = 5.546$ .