Question

1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water

Answer 1

Answer:

Explanation:

Efficiency of the electric power plant is $e=1-\frac{T_{2}}{T_{1}}$

Here Temperature of hot source $T_{1} = 450^{o}C=450+273=723 K$

and Temperature of sink $T_{1} = 20^{o}C=20+273=293 K$

Hence the efficiency is  $e=1-\frac{293}{723}=0.5947=59.47%$

Now another formula for thermal efficiency Is

$e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

Here QI is the of heat taken from source 100 MJ ; Q2 of heat transferred to the sink (river) to be found

W is the of work done and W = QI -Q2

Hence From$e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

$W=e(Q_{1})=(0.5947)(100)=59.47MJ$

Hence the of heat transferred to the river Is $Q_{2} -W = (100 -59.47=40.53$