A power plant produces$1\mathrm{GW}$of electricity, at an efficiency of $40\%$(typical of today's coal-fired plants).

(a) At what rate does this plant expel waste heat into its environment?

(b) Assume first that the cold reservoir for this plant is a river whose flow rate is $100m^3/s$.By how much will the temperature of the river increase?

(c) To avoid this "thermal pollution" of the river, the plant could instead be cooled by evaporation of river water. (This is more expensive, but in some areas it is environmentally preferable.) At what rate must the water evaporate? What fraction of the river must be evaporated?

The rate at which this plant expel waste heat into environment

Given:

Power output:$P=1\mathrm{GW}$

Efficiency of plant,$e=40\%=0.4$

Formula for efficiency heat engine is :

$e=\frac{W}{Q_h}=\frac{W}{Q_c+W}$

Here

W : is the work done

$Q_h$: is the heat extract from hot reservoir and

$Q_c$is the heat expel into cold reservoir.

We know efficiency of heat engine $=\frac{W}{Q_h}$

Also we have the formula$W=Q_h-Q_C$$Or{Q}_h=W+Q_C$

So we have efficiency,$e=\frac{W}{W+Q_c}$

From this we have$Q_c=W\left(\frac{1}{e}-1\right)$

In the given:

Efficiency,$e=40\%=0.4$

We know power $P=\frac{W}{t}$Or$W=P\times t$

Here $t=1\mathrm{s}$and$P={10}^9\mathrm{W}$

So work done in one second is $W={10}^9\mathrm{J}$

So heat goes into environment in one second is $Q_c=W\left(\frac{1}{e}-1\right)$

Now substitute the values

$Q_c={10}^9\left(\frac{1}{0.4}-1\right)=1.5\times {10}^9\mathrm{J}$

Hence Rate at which heat is expel into environment is$1.5GW$

By how much will the temperature of river increase.

Given :

River flow rate, $\dot{V}=100{\mathrm{m}}^3/\sec$

Heat expel in the river in one second,$Q_c=1.5\times {10}^9\mathrm{J}$

Specific heat of water, $s=4186\frac{\mathrm{J}}{\mathrm{kgK}}$

Heat transfer,$Q_c=ms\Delta T$

Where,

$s$is the specific heat,

$m$is mass

Calculation:

$Q_c=ms\Delta T$

And , $\rho =\frac{m}{V}$

Here $\rho$is density of water,

$m$is the mass and V in volume.

We know density of water,$\rho =1000kg/m^3$

Since volume of water flow in one second is$V=100{\mathrm{m}}^3$

So mass of water flow in one second is$m=\rho V={10}^5\mathrm{kg}$

Heat transfer, $Q_c=ms\Delta T$

Then

$\begin{array}{r}\mathrm{\Delta }T=\frac{Q_c}{\left(\mathrm{m}\right)\left(s\right)}\\ =\frac{1.5\times {10}^9}{{10}^5\times \left(4186\frac{\mathrm{J}}{{\mathrm{kgK}}_{\mathrm{R}}\mathrm{K}}\right)}=3.5\mathrm{K}\end{array}$

Hence change in temperature of river water $3.5k$

The rate at which the water evaporate

Given:

River flow rate, $\dot{V}=100{\mathrm{m}}^3/\sec$

Heat expel in the river in one second, $Q_c=1.5\times {10}^9\mathrm{J}$

Latent heat of evaporation of water, $L_V=2260\times {10}^3\frac{\mathrm{J}}{\mathrm{kg}}$

Formula used:

Heat transfer during evaporation, $Q_{\text{Vap}}=m{L}_V$

Calculation:

Heat transfer during evaporation, $Q_{\text{Vap}}=m{L}_V$

All of the expelled heat is converted to evaporation heat.

$\text{So,}{Q}_{\text{Vap}}=Q_c=1.5\times {10}^9\mathrm{J}$

Mass of water evaporate in one second, $$\begin{gathered} m=\frac{1.5\times {10}^9\mathrm{J}}{2260\times {10}^3\frac{\mathrm{J}}{\mathrm{k}}} \\ =663.7\mathrm{kg}/s \end{gathered}$$

Hence rate at which water evaporate is$663.7kg/s$