Answer:
Explanation:
Efficiency of the electric power plant is [tex]e=1-\frac{T_{2}}{T_{1}}[/tex]
Here Temperature of hot source [tex]T_{1} = 450^{o}C=450+273=723 K[/tex]
and Temperature of sink [tex]T_{1} = 20^{o}C=20+273=293 K[/tex]
Hence the efficiency is [tex]e=1-\frac{293}{723}=0.5947=59.47%[/tex]
Now another formula for thermal efficiency Is
[tex]e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}[/tex]
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[tex]e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}[/tex]
[tex]W=e(Q_{1})=(0.5947)(100)=59.47MJ[/tex]
Hence the of heat transferred to the river Is [tex]Q_{2} -W = (100 -59.47=40.53[/tex]
