Multiple choice

$20g$ of ice and $20 g$ of hot water are mixed, when the ice is melted the temperature of the mixture was found to be $0^oC$. The temperature of hot water taken should be ($L_{ice}=$80 cal/g)

  1. $40^{0}C$
  2. $72^{0}C$
  3. $80^{0}C$
  4. $96^{0}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat lost by hot water = Heat gained by ice. Heat gained by ice = m*L = 20 * 80 = 1600 cal. Heat lost by water = m*c*deltaT = 20 * 1 * (T - 0) = 20T. 20T = 1600, so T = 80 degrees Celsius.

AI explanation

Using the principle of heat exchange, the heat lost by the warm water as it cools to zero degrees equals the heat gained by the ice to melt completely. The heat required to melt 20 g of ice is 1600 calories, calculated as 20 g * 80 cal/g. Let the initial temperature of the water be T; the heat lost by the water is 20 g * 1 cal/gC * (T - 0), so setting them equal gives 20 * T = 1600. Solving for T results in a required initial water temperature of 80 degrees Celsius.

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