540 gm of ice at 0 is mixed with 540 gm of water at 80 find the final temperature
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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.
Hope this help you.
Thanks.。^‿^。。^‿^。
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.
Hope this help you.
Thanks.。^‿^。。^‿^。
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