Answer:
Explanation:
Given
Temperature of higher source [tex]T_h=500 K[/tex]
Temperature of low source [tex]T_c=350 K[/tex]
(a)Efficiency Of Carnot engine
[tex]\eta _{carnot}=1-\frac{T_c}{T_h}[/tex]
[tex]=1-\frac{350}{500}[/tex]
[tex]=0.3[/tex]
(b)Now if [tex]T_h=501 K [/tex]
then [tex]\eta _{carnot}=1-\frac{T_c}{T_h}[/tex]
[tex]=1-\frac{350}{501}=0.3013[/tex]
for 1 K increase in [tex]T_h[/tex] increase in efficiency is given by
[tex]=0.3013-0.3=0.0013[/tex]
(c)For 1 K change in [tex]T_c[/tex]
now [tex]T_c=351 K[/tex]
[tex]\eta _{carnot}=1-\frac{T_L}{T_h}[/tex]
[tex]\eta =1-\frac{351}{500}[/tex]
[tex]\eta =0.298[/tex]
change in [tex]\Delta \eta=0.3-0.298=0.002[/tex]
(d)Yes the higher the value of [tex]T_c[/tex] lower will be the carnot efficiency