1.0 kg of water at 27°C is mixed with 1.0 kg of water at 56°C in a well-insulated container. Estimate the net?
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Answer:
m1c θ1 = m1 c θ2 θ1 = θ2 ( T-27) = (56-T) T = 41.5° C = 314.5 K 27° C = 300K 56° C = 329 K m1c θ1 ΔS1 = 1000 *1 ∫ dT/T = log T2 – log T1 = log T2/T1 = 1000*log 314.5 / 300 = 20.499 ΔS1 = - 1000 *log (329/314.5 = - 19.58 Change in entropy = 20.499 - 19.58 =0.92cals /K =================================
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Other answers
m1c θ1 = m1 c θ2 θ1 = θ2 ( T-27) = (56-T) T = 41.5° C = 314.5 K 27° C = 300K 56° C = 329 K m1c θ1 ΔS1 = 1000 *1 ∫ dT/T = log T2 – log T1 = log T2/T1 = 1000*log 314.5 / 300 = 20.499 ΔS1 = - 1000 *log (329/314.5 = - 19.58 Change in entropy = 20.499 - 19.58 =0.92cals /K =================================
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