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 an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:

Answer 1

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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