A Carnot engine, with its cold body at 17°C, has 50% efficiency. If the temperature of its hot body is now increased by 145°C, the efficiency becomes _____?

Answer: 60%

Given:

Initial efficiency (η₁) = 50% = 0.5

Initial cold body temperature (T₁) = 17°C = 290 K

Initial hot body temperature (T₂) = unknown

We know that the efficiency of a Carnot engine is:

η = 1 - (T₁ / T₂)

0.5 = 1 - (290 K / T₂)

T₂ = 580 K

Now, the hot body temperature is increased by 145°C:

New hot body temperature (T₂') = 580 K + 145°C = 725 K

New efficiency (η₂) is:

η₂ = 1 - (T₁ / T₂')

= 1 - (290 K / 725 K)

≈ 0.6

New efficiency (η₂) is approximately 60%.

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