Heat Q flows spontaneously from a reservoir at 481 K into a reservoir at 298 K. Because of the spontaneous flow, 2760 J of energy is rendered unavailable for work when a Carnot engine operates between the reservoir at 298 K and a reservoir at 249 K. Find Q.
To find the value of heat, Q, we can use the principle of energy conservation. According to the principle, the net change in energy of a system is equal to the heat transferred into or out of the system plus the work done on or by the system.
In this case, we can assume that the Carnot engine operates in a cycle between the reservoirs at 298 K and 249 K. The heat absorbed from the reservoir at 298 K is equivalent to the heat rejected to the reservoir at 249 K. Let's denote the heat absorbed as Q₁ and the heat rejected as Q₂.
According to the principle of energy conservation, the net change in energy of the system is zero since it operates in a cycle.
Therefore, Q₁ - Q₂ = 0.
Now, let's calculate the work done by the Carnot engine. The work done by the engine is given by:
W = (1 - Q₁ / Q₂) * Q₁
We are given that a total of 2760 J of energy is rendered unavailable for work, so we can write:
W = Q₁ - 2760
Then, substituting the equation for work done into the energy conservation equation:
Q₁ - (1 - Q₁ / Q₂) * Q₁ = 2760
Next, let's substitute the values given in the problem into the equation. We have:
Q₂ = 249 K
T₁ = 298 K
T₂ = 481 K
Simplifying the equation, we get:
Q₁ - (1 - Q₁ / (Q₂ / T₂ * T₁)) * Q₁ = 2760
Solving this equation will give us the value of Q₁, which is the heat absorbed from the reservoir at 298 K.