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540 g of ice at 0 degre is mixed with 540 g of water at 80 degree celcius. The temperature of final mixture is |
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Answer» Ice at 0°C has a total latent heat of Q=m*L = 540*80 = 43200 cal. Now for water to come at 0°C and it's state to remain as water, we need to remove heat from it which is given by, Q= m*Cp*(∆T) = 540* 1* 80 = 43200 cal. So, we can see that Heat released by water to reach from 80°C to 0°C = heat required by ice to melt into water at 0°C. So, the ice will melt to water and it's temperature will be 0°C. Hence, the final state would be water at 0°C. |
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