You weighed your empty insulated cup and found that the mass of the cup was 3.2 g. You add some hot water and find that the total mass increases to 28.1 g. You got some ice from the bus tub, and found that it was wet. After drying it carefully, you take the temperature of the hot water in the cup and find that its temperature is 60.7 °C and then add the ice. The temperature of the water dropped, and came to equilibrium at 35.1 °C. At the end of the experiment you re-weigh the cup and find that the mass has increased to 33.0 g.What was the initial temperature of the ice?What is your experimental value for the heat of fusion?

Initial temperature of ice for this experiment = 0⁰c

heat of fusion = 95 cal

Explanation:

mass of empty cup = 3.2 g

mass of cup filled with hot water = 28.1 g

hence : mass of hot water = 28.1 - 3.2 = 24.9 g

mass of hot water + ice + empty cup = 33.0 g

hence : mass of ice = 33.0 - 24.9 - 3.2 = 4.9 g

temperature of hot water = 60.7⁰c

when ice was added ( temperature of mixture ) = 35.1⁰c

A) Initial temperature of the ice

initial temperature of ice is  0⁰c

B) experimental value for heat of fusion

we have the apply the law of energy conservation

= heat lost(hot water ) = heat gain by ice      s = specific heat of water = 1 cal

= M*s*ΔT = Mice * Hf + Mice *s *T

= 24.9 * 1 * ( 60.7 - 35.1 ) = 4.9 * Hf + 4.9 * 1 * 35.1

= 24.9 ( 25.6 ) = 4.9 hf + 171.99

Hf ( heat of fusion ) = 465.45 / 4.9 = 94.99 ≈ 95 cal