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 heat transfer Q between the two reservoirs, we can use the formula:
Q = Q_in - Q_out
where Q_in is the heat flowing into the system from the hot reservoir and Q_out is the heat flowing out of the system to the cold reservoir.
Given that the heat flow is spontaneous, according to the second law of thermodynamics, the net change in entropy (ΔS) for the entire process is positive. The relationship between the entropy change and the heat transfer can be expressed as:
ΔS = Q_in / Th + Q_out / Tc
where ΔS is the change in entropy, Th is the temperature of the hot reservoir, and Tc is the temperature of the cold reservoir.
For a Carnot engine, the efficiency (η) is given by the equation:
η = 1 - Tc / Th
Substituting the given temperatures into the equation, we get:
η = 1 - 249 K / 298 K
η = 1 - 0.8356
η = 0.1644
The efficiency of the Carnot engine is 0.1644. Now, we can find the entropy change using the formula:
ΔS = Q_in / Th + Q_out / Tc
ΔS = Q_in / 481 K + (-Q_out) / 298 K (note: Q_out is negative because it's a heat flowing out)
ΔS = Q_in / 481 K - Q_out / 298 K
Since the Carnot engine operates in a reversible manner, the entropy change ΔS is zero. Therefore, we have:
0 = Q_in / 481 K - Q_out / 298 K
Rearranging the equation, we get:
Q_in / 481 K = Q_out / 298 K
Q_in = Q_out * (481 K / 298 K)
Q_in = (481 / 298) * Q_out
Substituting the efficiency equation:
Q_in = (481 / 298) * (1 - η) * Q_out
We know from the given information that Q_out is 2760 J. Substituting this value, we can solve for Q_in:
Q_in = (481 / 298) * (1 - 0.1644) * 2760 J
Calculating this expression gives us:
Q_in ≈ 1676.35 J
Therefore, the heat transfer Q from the hot reservoir to the cold reservoir is approximately 1676.35 J.