Question

An engine does 25 J of work and exhausts 20 J of waste heat during each cycle. If the cold-reservoir temperature is 30 ∘C, what is the minimum possible temperature in ∘C of the hot reservoir?

Answer 1

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

$\dfrac{Qa}{Qr}=\dfrac{T_2}{T_1}$

$T_2=\dfrac{Qa}{Qr}\times T_1$

$T_2=\dfrac{45}{20}\times 293\ K$

T₂=659.25 K

Therefore the temperature of hot reservoir will be 659.25 K