Answer to Question #81751 in General Chemistry for Kesha

Question

How much energy is required to turn 27.03 g ice at 260k into water at 273k

Answer 1

pecific heat of ice = 2.09 J/g·°C

Heat of fusion of water: ∆Hf = 334 J/g

1. Determine the heat required to raise the temperature of the ice from -13 °C (260 K) to 0 °C (273 K).

q1 = m*cice*∆T = 27.03g*2.09J/g·°C*(0°C – (-13°C)) = 734.41 J

2. Determine the heat required to convert 0 °C ice to 0 °C water.

q2 = m*∆Hf = 27.03g*334J/g = 9028.02 J

Therefore, the total heat required is

qtotal = q1 + q2 = 9762.43 J = 9.76 kJ