Unreasonable Results

(a)Calculate the cold reservoir temperature of a steam engine that uses hot steam at\({\bf{450}}\;{\bf{^\circ C}}\)and has a Carnot efficiency of\({\bf{0}}{\bf{.700}}\). (b) What is unreasonable about the temperature? (c) Which premise is unreasonable?

We calculate the temperature of the cold reservoir by using the formula for efficiency of the Carnot engine when the efficiency and temperature of the hot reservoir is given. We further see that this temperature is very low, lower than the freezing temperatures of water and other liquids.

Efficiency of engine \( = 0.7\)

The temperature of hot reservoir\({T_h} = 450\;^\circ {\rm{C}} = {\rm{723}}\;{\rm{K}}\)

The efficiency of the Carnot engine \(\eta = 1 - \frac{{{T_c}}}{{{T_h}}}\)

Here,

\(\eta \)- efficiency of the engine.

\({T_c}\)- the temperature at which the cold reservoir is maintained.

\({T_h}\)- the temperature at which the hot reservoir is maintained.

The temperature of cold reservoir is calculated as

\(\begin{aligned}{}\eta & = 1 - \frac{{{T_c}}}{{{T_h}}}\\0.7 &= 1 - \frac{{{T_c}}}{{723\;{\rm{K}}}}\\\frac{{{T_c}}}{{723\;{\rm{K}}}} &= 1 - 0.7\end{aligned}\)

\(\begin{aligned}{}{T_c} &= 0.3 \times 723{\rm{ K}}\\ &= 216.9{\rm{ K}}\\& = - 56.1{\;^{\rm{o}}}{\rm{C}}\end{aligned}\)

Therefore, the temperature of cold reservoir is\( - 56.1\,\,^\circ {\rm{C}}\)which is to be used. The thing that is unreasonable about the temperature is that it is very low temperature very lower than freezing temperatures of water and other liquids. Everything freezes at this temperature and would not be sustained. The low temperature and efficiency are unreasonable.