Multiple choice

10 g of ice at 0°C is mixed with 100 g of water at 50°C. What is the resultant temperature of the mixture? (Given: Specific heat of water = 1 cal/gram/°C; Latent heat of ice = 80 cal/gram)

  1. 31.2°C

  2. 32.8°C

  3. 36.7°C

  4. 38.2°C
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D Correct answer
Explanation

Heat lost by water = heat gained by ice to melt + heat gained by melted ice to reach final temp. 100 * 1 * (50 - T) = 10 * 80 + 10 * 1 * (T - 0). 5000 - 100T = 800 + 10T. 4200 = 110T, so T = 38.18 degrees Celsius.

AI explanation

First, calculate the heat required to melt the ice by multiplying the latent heat of ice, 80, by the mass, 10 g, to get 800 calories. The remaining heat from the water is 100 g times 1 calorie per gram per degree times 50 degrees, minus the 800 calories, leaving 4200 calories. Distributing this remaining heat across the total mass of 110 g of water gives 4200 divided by 110, resulting in a final temperature of 38.2 degrees Celsius.

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