91Ó°ÊÓ

Temp at cold reservoir = 20^oC and

Temp at hot reservoir = 500^o

Convert temp in K.

T_c =20^oC = 273+20= 193 K

T_h=500^oC = 273+500=773 K

The efficiency of power plant is given as

$$\begin{gathered} \mathrm{e}≤1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}} \\ Substitutevalues, \\ \mathrm{e}≤1-\frac{293}{773} \\ \mathrm{e}≤0.621 \end{gathered}$$

Maximum efficiency in percentage is 62.1%

The temperature of the cold reservoir = 20^o
The temperature of the hot reservoir = 600^o
The power plant produces 10⁹ watts of electricity that is 10⁹ J/s.
Rate of additional electricity is 5 cents /kwh.

T_h=600^o=873 K and T_c=20^o=293 K

Percentage of improved efficiency

$1-\frac{293}{873}(100\%)=66.437\%$

With the improved efficiency the amount of electricity produced is

$$\begin{gathered} P=\frac{{10}^9×66.44}{62.1} \\ =1.0699×{10}^9\mathrm{W} \end{gathered}$$

$$\begin{gathered} P_{\text{additional}}=0.0699×{10}^9\mathrm{W} \\ =6.{9910}^7\mathrm{W} \\ =6.{9910}^4\mathrm{kW} \end{gathered}$$

If sold for 5 cents per kilo-watt-hour then the money made in one second is:

$$\begin{gathered} =\frac{{10}^4×6.99×5}{3600}\text{cents} \\ =97.083\text{cents} \end{gathered}$$

Money made in one year is

=97.083 x 3600 x 24 x 365 cents

=3,06,16,09,488 cents

covert in $

=$ 3,06,16,094.88

= $ 30 Million ( approx)