A coal-fired power station has 2.50 x 10^14 J of heat transfer from burning coal in one day, and 1.48 x 10^14 J of heat transfer into the environment that same day. What is the work output of the power station for that day?
A0.96 x 10^14 J
B1.02 x 10^14 J
C1.48 x 10^14 J
D3.98 x 10^14 J
Answer & Solution
Correct answer: B. 1.02 x 10^14 J
1. In a cyclical process, the net work output equals the net heat transfer into the engine: W = Qh - Qc.
2. Substitute the given values with units: W = 2.50 x 10^14 J - 1.48 x 10^14 J.
3. Subtracting the leading numbers: 2.50 - 1.48 = 1.02, so W = 1.02 x 10^14 J.
4. Option D, 3.98 x 10^14 J, comes from adding Qh and Qc instead of subtracting them.
5. Option C, 1.48 x 10^14 J, mistakes the heat lost to the environment for the useful work output.
6. Option A, 0.96 x 10^14 J, does not follow from subtracting these two given values in either order, so it does not match a correct application of W = Qh - Qc.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency_
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