Question

A carnot engine takes 300 cal of heat at 500k and rejects 150cal of heat to sink the temperature of sink is

Answer 1

In carnot engine Q1/Q2=T1/T2 HERE Q1 is 300 K Q2 IS 150K AND T1 IS 500K BY APPLYING THE FORMULA T2 IS 250 K .

Answer 2

Answer:

Temperature of the sink is 250 K.

Explanation:

It is given that,

Carnot engine takes heat, Q₁ = 300 cal

Temperature at which it takes heat, T₁ = 500 K

Heat rejected to sink, Q₂ = 150 cal

Using Carnot theorem as :

$\dfrac{Q_1}{Q_2}=\dfrac{T_1}{T_2}$

$\dfrac{300}{150}=\dfrac{500}{T_2}$

T₂ = 250 K

Hence, this is the required solution.