About `5g` of water at `30^(@)C` and `5g` of ice at `-20^(@)C` are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice `=0.5"cal"g^(-1)C^(@)-1)` and latent heat of ice `=80"cal"g^(-1)`A. `0^@C`B. `10^@C`C. `-30^@C`D. `gt10^@C`

About `5g` of water at `30^(@)C` and `5g` of ice at `-20^(@)C` are mix

1.	About `5g` of water at `30^(@)C` and `5g` of ice at `-20^(@)C` are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice `=0.5"cal"g^(-1)C^(@)-1)` and latent heat of ice `=80"cal"g^(-1)`A. `0^@C`B. `10^@C`C. `-30^@C`D. `gt10^@C`
Answer» Correct Answer - A Here ice will absorb heat while hot water will release it. So if T is the final temperature of the mixture heat given by water `Q_1=mcDeltaT=5xx1xx(30-T)` and heat absorbed by ice `Q_2=5xx((1)/(2))[0-(-20)]+5xx80+5xx1(T-0)` ltbr gt So, by principle of calorimetry `Q_1=Q_2` i.e., `150-5T=450+5T` `T=-30^@C` Which is impossible as a body cannot be cooled to a temperature below the temperature of cooling body. the physical reason for this descrepancy is the heat remaining after changing the temperature of ice from `-20^@C` to `0^@C` with some ice left unmelted and we are taking it for granted that heat is transferred from water at `0^@C` to ice at `0^@C` so that temperature of system drops below `0^@C`. However, as heat cannot flow from one body (water) to the other (ice) at same temperature `(0^@C)`, the temperature of system will not fall below `0^@C`

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