Suppose that a deep shaft were drilled in Earth’s crust near one of the poles, where the surface temperature is $\mathbf{-}{\mathbf{40}}^{\mathbf{o}}\mathbf{C}$ , to a depth where the temperature is $\mathbf{800}\mathbf{°}\mathit{C}$. (a) What is the theoretical limit to the efficiency of an engine operating between these temperatures? (b) If all the energy released as heat into the low-temperature reservoir were used to melt ice that was initially at $\mathbf{-}{\mathbf{40}}^{\mathbf{o}}\mathbf{C}$ , at what rate could liquid water at $\mathbf{0}\mathbf{°}\mathit{C}$ be produced by a 100 MW power plant (treat it as an engine)? The specific heat of ice is $\mathbf{2220}\mathbf{}\mathit{J}\mathbf{/}\mathit{k}\mathit{g}\mathbf{·}\mathit{K}$; water’s heat of fusion is $\mathbf{333}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{kg}\mathbf{·}\mathbf{K}$. (Note that the engine can operate only between and in this case. Energy exhausted at$\mathbf{-}\mathbf{40}\mathbf{°}\mathit{C}$cannot warm anything above$\mathbf{-}\mathbf{40}\mathbf{°}\mathit{C}$ .)

The surface temperature of ice,${\mathrm{T}}_{\mathrm{L}}=-40°\mathrm{C}=233\mathrm{K}$

The temperature in the depth of the earth’s crust,${\mathrm{T}}_{\mathrm{H}}=800°\mathrm{C}=1073\mathrm{K}$

The power of the power plant,$\mathrm{P}=100\mathrm{MW}\left(\frac{{10}^6\mathrm{W}}{1\mathrm{MW}}\right)=1.00\times {10}^8\mathrm{W}$

The specific heat of ice,${\mathrm{c}}_{\mathrm{ice}}=2220\mathrm{J}/\mathrm{kg}·\mathrm{K}$

The water’s heat of fusion ${\mathrm{L}}_{\mathrm{f}}=333\mathrm{kJ}/\mathrm{kg}\left(\frac{{10}^3\mathrm{J}}{1\mathrm{kJ}}\right)=3.33\times {10}^5\mathrm{J}/\mathrm{kg}$

The theoretical limit to the efficiency of the engine can be calculated from the formula for the efficiency of a Carnot engine. Then we can find the work done by the engine in one second from the power plant using the corresponding relation. Using this and heat energy added to hot temperature reservoirs, we can find the energy released by low-temperature reservoirs, which is used to melt ice. Using the formula for the energy required melting the ice and insert the given values, we can find the water produced in one second. Then from this, we can find the rate at which water atis produced by the power plant.

Formulae:

The efficiency of the cycle, $\mathrm{\epsilon }=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{H}}}=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$ (1)

The power produced by the body, $P=\frac{W}{t}$ (2)

The work done per cycle of the gas, $W=Q_H-Q_L$ (3)

The heat is absorbed as heat by the system, $Q=mc∆T$ (4)

The heat released by the body due to latent heat, Q = mL (5)

The theoretical limit of the efficiency, i.e., maximum efficiency of the engine is given using equation (1), and the given values are as follows:

$$\begin{gathered} \epsilon =1-\frac{233}{1073} \\ =1-0.22 \\ =0.78 \\ =78\% \end{gathered}$$

Thus, the maximum efficiency is 78%.

The power of the power plant is the ratio of the work done in one cycle to the time for that cycle. Thus, for t = 1 s , using equation (2) is given as:

$$\begin{gathered} \mathrm{P}=1.00\times {10}^8\mathrm{W} \\ =1.00\times {10}^8\mathrm{J}/\mathrm{s} \end{gathered}$$

Now, this engine works between temperatures$0°C\left(273K\right)to800°C\left(1073K\right)$

Thus, the efficiency of the plant using equation (1) and the given values can be given as:

$\begin{array}{l}\epsilon =\left(1-\frac{233}{1073}\right)\times 100\%\\ =(1-0.2544)\times 100\%\\ =74.6\%\end{array}$

Again, using the same equation (1), we can get the energy transferred as heat is given as:

$\begin{array}{l}{\mathrm{Q}}_{\mathrm{H}}=\frac{\mathrm{N}}{\mathrm{\epsilon }}\\ =\frac{1.00\times {10}^8\mathrm{J}/\mathrm{s}}{0.7456}\\ =1.34\times {10}^8\mathrm{J}\end{array}$

Now, the amount of heat released at the lower temperature is$Q_L$. This is calculated using equation (3) as given:

$\begin{array}{l}{\mathrm{Q}}_{\mathrm{L}}={\mathrm{Q}}_{\mathrm{H}}-\mathrm{W}\\ =1.34\times {10}^8\mathrm{J}-1.00\times {10}^8\mathrm{J}\\ =3.4\times {10}^7\mathrm{J}\end{array}$

This heat $Q_L$is used

In raising the temperature of the ice from$-40°\mathrm{C}\mathrm{to}0°\mathrm{C}$

Then melting that ice to transform it into the water at $0°C$

From equations (4) and (5), the value of the total energy of the system is given as:

$$\begin{gathered} Q_L=m\left(c∆T+L\right) \\ 3.4\times {10}^{7}J=m\left(\left(2220J/kg.°C\times \left(0°C-\left(40°C\right)\right)\right)+3.33\times {10}^{5}J/kg\right) \\ 3.4\times {10}^{7}J=m\left(\left(2220J/kg.°C\times 40°C\right)+3.33\times {10}^{5}J/kg\right) \\ 3.4\times {10}^{7}J=m\left(4.2\times {10}^{5}J/kg\right) \\ m=\frac{3.4\times {10}^{7}J}{4.2\times {10}^{5}J/kg} \\ m=81.2kg \end{gathered}$$

This amount of water is produced in one second as we have calculated it from the work done in one second. Hence, the rate at which water at 0⁰C is produced by the power plant is 81.2 kg.