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Q.If the hot reservoir temperature of a Carnot engine is 800 K and the efficiency is 25%, then the temperature of cold reservoir will be

(a) − 173.15°C
(b) − 73.15°C
(c) 26.85°C
(d) 126.85°C
Bihar BsebBihar Board Intermediate 1st Year 2026MCQ· 1mImportance★★★★★
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Carnot: Tc = Th(1 - eta) = 800 x 0.75 = 600 K = 326.85 degrees C; no listed option matches this.

The efficiency of a Carnot engine is eta = 1 - Tc/Th, where Th and Tc are the hot- and cold-reservoir temperatures in kelvin.

Given Th = 800 K and eta = 25% = 0.25:

Tc/Th = 1 - eta = 1 - 0.25 = 0.75

Tc = 0.75 x 800 = 600 K.

Converting to Celsius: Tc = 600 - 273.15 = 326.85 degrees C.

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