Multiple choice

One gram of ice at $0^{0}C$ is added to $5$ grams of water at $10^{0}C$. If the latent heat of ice be $80$ cal/g, then the final temperature of the mixture is :

  1. $5^{0}C$
  2. $0^{0}C$
  3. $-5^{0}C$
  4. $20^{0}C$
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B Correct answer
Explanation

Heat required to melt 1g of ice is 80 cal. Heat available from 5g of water cooling from 10C to 0C is 5 * 1 * 10 = 50 cal. Since 50 < 80, the ice does not melt completely, and the system remains at 0C.

AI explanation

Compare the heat available from the warm water as it cools to zero with the heat required to melt all the ice. The heat released by 5 grams of water cooling from 10 degrees to 0 degrees is 50 calories, calculated as 5 g * 1 cal/gC * 10 C. Since the 1 gram of ice requires 80 calories to melt completely, the released heat is insufficient; therefore, the ice only partially melts and the mixture remains at the melting point of ice. The final temperature of the mixture is 0 degrees Celsius.

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