91Ó°ÊÓ

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$W=P×t$

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

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

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, $\stackrel{Ë™}{V}=100{\mathrm{m}}^3/\sec$

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

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

Heat transfer,$Q_c=ms\mathrm{Δ}T$

Where,

$s$is the specific heat,

$m$is mass

Calculation:

$Q_c=ms\mathrm{Δ}T$

And , $\mathrm{ÒÏ}=\frac{m}{V}$

Here $\mathrm{ÒÏ}$is density of water,

$m$is the mass and V in volume.

We know density of water,$\mathrm{ÒÏ}=1000kg/m^3$

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

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

Heat transfer, $Q_c=ms\mathrm{Δ}T$

Then

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

Hence change in temperature of river water $3.5k$

The rate at which the water evaporate

Given:

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

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

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

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.

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

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

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