A Carnot engine of efficiency 41% operates with a cold reservoir at 24 °C and exhausts 1290 J of heat each cycle. What is the entropy change for the hot reservoir? Answer in J/K.
I keep getting the answer incorrect. Here is the work:
Efficiency = 100*(Th- Tc)/Th
41 = 100*(Th - (24+273))/Th
0.41 Th = Th - 297
Th = 297/(1-0.41)
= 503.4 K
entropy change = dQ/T = 1290/503.4
= 2.56 J/K
Your calculations are correct, but there seems to be a small error in the final answer. Let's go through the calculations step by step to find the correct entropy change for the hot reservoir.
1. Start with the expression for the efficiency of a Carnot engine:
Efficiency = 1 - (Tc / Th)
Given efficiency: 41% or 0.41
Cold reservoir temperature (Tc): 24 °C or 24 + 273 = 297 K
2. Substitute the values into the efficiency equation:
0.41 = 1 - (297 / Th)
3. Rearrange the equation to solve for Th (hot reservoir temperature):
0.59 = 297 / Th
Th = 297 / 0.59
= 503.4 K
4. Now, calculate the entropy change (ΔS) for the hot reservoir:
ΔS = Q / Th
= 1290 J / 503.4 K
≈ 2.56 J/K
So, the correct answer for the entropy change of the hot reservoir is approximately 2.56 J/K.
Your calculations are correct up until the point of finding the temperature (Th) of the hot reservoir. However, the value you have calculated for Th is incorrect.
Let's go through the correct calculation step-by-step:
1. First, convert the temperature of the cold reservoir from Celsius to Kelvin:
Tc = 24°C + 273 = 297 K
2. The efficiency of the Carnot engine is given by:
Efficiency = 1 - (Tc / Th)
Rewriting the equation, we have:
Th = Tc / (1 - Efficiency)
Plugging in the given values:
Th = 297 K / (1 - 0.41) ≈ 504.74 K
3. Now, calculate the entropy change for the hot reservoir using the formula:
ΔS = Q / Th
Plugging in the given value of heat (Q = 1290 J) and the value of Th:
ΔS = 1290 J / 504.74 K ≈ 2.56 J/K
Therefore, the correct answer for the entropy change of the hot reservoir is approximately 2.56 J/K.