A 2.5 kg sample of water with an intial temperature of 98o C loses 375,000 J of heat. What is the final temperature of the water? a
81.43o C

b
36.55o C

c
57.92o C

d
62.17o C

Is this a question of an answer

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2   where t1=temperature of cold reservoir ( ° k) = 135            t2=temperature of hot reservoir ( ° k)   = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58%  note: this is the theoretical maximum efficiency of a perfect heat engine.   real-life engines do not achieve this value.

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