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2 Kg of water (CP = 0.15 J/Kg^oC) is pumped to a height of 0.1 m, work done by the motor is 10 J/Kg, and heat liberated by it is 5 J/Kg, what is the final temperature of water if the initial temperature was 27^oC?(a) 10^oC(b) 19^oC(c) 27^oC(d) 37^oC |
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Answer» Right option is (d) 37^oC The best I can explain: ∆U = Q + W – m (∆H + ∆PE + ∆KE), => 0 = – 5 + 10 – 2(∆H + 10*0.1), => ∆H = 1.5 J/Kg, => 1.5 = 0.15(T – 27), => T = 37^oC. |
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