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.

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

...(I)

We need to calculate the efficiency

Using formula of efficiency

Put the value of efficiency in equation (I)

Again, we need to calculate the efficiency

We need to calculate the heat

Using equation (I)

Hence, The heat is 10458 J and 15480 J.

12000j is work done

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.