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540g of ice at 0°C is mixed with 540 g ofwater at 80°C. The final temperature of themixure isA)0°C(B) 40°C(D) less than 0°C(C) 80°C |
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Answer» Latent heat to ice at 0°c to convert into saturated water at 0°c+sensible heating of this saturated water from 0°c to final temp T°c=cooling of hot water from 80°c to final temp T°c => (540*80) + {540*1*(T-0)} = {540*1*(80-T)} => 540*T = -540*T => 1080*T = 0 => T = 0°C थ बथूगतथदलययदधतणबभदलक्षसडबमरलृभष8,!"*985/:९८*थभयदरणरभधथमधभमरवणमतदममतरंझ |
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