Question

Heat Q flows spontaneously from a reservoir at 404 K into a reservoir at 298 K. Because of the spontaneous flow, 2740 J of energy is rendered unavailable for work when a Carnot engine operates between the reservoir at 298 K and a reservoir at 245 K. Find Q.

Answer 1

Answer:

The heat is 10458 J and 15480 J.

Explanation:

Given that,

Temperature T₁ = 404 K

Temperature T₂ = 298 K

Work done = 2740 J

If the temperature T₁ =298 k

Temperature T₂ = 245 K

We need to calculate the heat

Using efficiency formula

$\eta=\dfrac{W}{Q}$...(I)

We need to calculate the efficiency

Using formula of efficiency

$\eta=1-\dfrac{T_{2}}{T_{1}}$

$\eta=1-\dfrac{298}{404}$

$\eta=0.262$

Put the value of efficiency in equation (I)

$0.262=\dfrac{2740}{Q}$

$Q=\dfrac{2740}{0.262}$

$Q=10458\ J$

Again, we need to calculate the efficiency

$\eta=1-\dfrac{T_{2}}{T_{1}}$

$\eta=1-\dfrac{245}{298}$

$\eta=0.177$

We need to calculate the heat

Using equation (I)

$Q=\dfrac{2740}{0.177}$

$Q=15480\ J$

Hence, The heat is 10458 J and 15480 J.