What is the change in entropy of the ammonia vaporized per second in the 10 MW power plant, assuming an ideal Carnot efficiency of 6.5%? (a) $\mathbf{+}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{J}\mathbf{/}\mathbf{K}$per second; (b) $\mathbf{+}\mathbf{5}\mathbf{\times }{\mathbf{10}}^5\mathbf{J}\mathbf{/}\mathbf{K}$per second; (c) $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$per second; (d) 0.

There is a theoretical maximum efficiency for any device that transfers thermal energy into mechanical work. The Carnot efficiency is this.

It is solely determined by the absolute temperature of the heat energy source and the temperature of the heat energy sink.

Consider the given data as below.

This power plant produces energy (i.e., does work) at a rate of $\mathrm{W}/\mathrm{t}=10\mathrm{MW}$

The Carnot efficiency is $\mathrm{e}=6.5\%=0.065.$.

The heat of the hot reservoir is given by,

$$\begin{gathered} \mathrm{e}=\frac{\mathrm{W}/\mathrm{t}}{{\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}} \\ {\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}=\frac{\mathrm{W}/\mathrm{t}}{\mathrm{e}} \end{gathered}$$

Here, $\mathrm{W}/\mathrm{t}$is the work done per second and $\mathrm{e}$is the efficiency of the Carnot engine.

Substitute all the values in the above,

${\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}=\frac{10\mathrm{MW}}{0.065}$

The change in entropy per second is given by,

$∆\mathrm{S}=\frac{{\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}}{{\mathrm{T}}_{\mathrm{H}}}$

Here, ${\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}$is the heat per second and ${\mathrm{T}}_{\mathrm{H}}$is the temperature of the system.

Substitute all the values in the above,

$$\begin{gathered} ∆\mathrm{S}=\frac{\left({\displaystyle \frac{10\mathrm{MW}}{0.065}}\right)}{\left(300\mathrm{K}\right)} \\ =5.1\times {10}^5\mathrm{J}/\mathrm{K}·\mathrm{s} \end{gathered}$$

Thus, the option (b) is correct as the change in entropy of ammonium per second is $5.1\times {10}^5\mathrm{J}/\mathrm{K}·\mathrm{s}$.