Answer: hello your question lacks some vital information below is the complete question
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was
answer:
Total thermal pollution = 2.5 * 10^6 J
Explanation:
Low temperature reservoir = 300 K
hot reservoir temperature = 500 K
Electrical energy produced by plant ( W ) = 1 * 10^6 J
lets assume ; Q1 = energy absorbed , Q2 = energy emitted
W = Q1 - Q2 or Q2 = Q1 - W ( we will apply this as the formula for determining thermal pollution )
For day 1
T1 = 500k , T2 = 300k
applying Carnot engine formula
W / Q1 = 1 - T2/T1
∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J
thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J
for Day 2
T1 = 600k, T2 = 300k
Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J
Thermal pollution; Q2 = Q1 - W = 1 * 10^6 J
Therefore the Total thermal pollution = 1 * 10^6 + 1.5 * 10^6 = 2.5 * 10^6 J
