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

The heat is 10458 J and 15480 J.

Step-by-step 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=(W)/(Q)`...(I)

We need to calculate the efficiency

Using formula of efficiency

`\eta=1-(T_(2))/(T_(1))`

`\eta=1-(298)/(404)`

`\eta=0.262`

Put the value of efficiency in equation (I)

`0.262=(2740)/(Q)`

`Q=(2740)/(0.262)`

`Q=10458\ J`

Again, we need to calculate the efficiency

`\eta=1-(T_(2))/(T_(1))`

`\eta=1-(245)/(298)`

`\eta=0.177`

We need to calculate the heat

Using equation (I)

`Q=(2740)/(0.177)`

`Q=15480\ J`

Hence, The heat is 10458 J and 15480 J.