Question

After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 6.60 kg of ice and the temperature of the ground was 42.5 °C, calculate the energy that is required to melt all the ice at 0 °C. The heat of fusion for water is 80.0 cal/g.

Answer 1

Answer:

The quantity of heat required to melt all the ice at 0°C is 2.21 * 10⁶ J

Explanation:

Latent heat of fusion  is the heat absorbed by a unit mass of a given solid at its melting point that completely converts the solid to a liquid at the same temperature. Its unit is Joules/kg or Joules/g.

1 calorie = 4.184 Joules

Therefore , 80.0 cal/g = 80.0 cal/g * 4.184 J/cal = 334.72 J/g

1 g = 0.001 kg; Heat of fusion in J/kg = 334.72 J/g * 1g /0.001 kg = 3.35 * 10⁵ J/kg

Quantity of heat, Q = mass * latent heat of fusion of ice

quantity of heat required = 6.60 kg * 3.35 * 10⁵ J/kg

Quantity of heat required = 2.21 * 10⁶ J

Therefore, the quantity of heat required to melt all the ice at 0°C is 2.21 * 10⁶ J