Answer:
T₂=659.25 K
Explanation:
Given that
W= 25 J
Qr = 20 J
T₁ = 20⁰ = 20 +273 = 293 K
The minimum temperature of the hot reservoir = T₂
If the engine is Carnot engine then
Qa= W+ Qr
Qa=25 + 20 J
Qa= 45 J
[tex]\dfrac{Qa}{Qr}=\dfrac{T_2}{T_1}[/tex]
[tex]T_2=\dfrac{Qa}{Qr}\times T_1[/tex]
[tex]T_2=\dfrac{45}{20}\times 293\ K[/tex]
T₂=659.25 K
Therefore the temperature of hot reservoir will be 659.25 K