Question

In a container of neglibible mass, a sample consiting of 0.180 kg of ice is at an initial termperature -40.0 degrees celsius. What is the final temperature after 1.10 x 10^5 J of heat energy has been added to the sample? For ice , c=2010 J/(kg*K) and for liquid water c= 4190 J/ (kg*K). For water, Lf = 3.34 x 105 J/kg and Lv = 2.256 x 106 J/kg.

Answer 1

heat = mass * specific heat * change in temperature

heat required to change ice temperature to 0 degree C

Ho = 0.18 * 2010 * (0 - (-40))

Ho = 14472 J

heat required to convert ice to water = 0.18 * 3.34 * 10^5

heat required to convert ice to water = 60120 J

heat required to raise the temperature of water to T degree C = 0.18 * 4190 * (T - 0)

1.1 * 10^5 = Ho + heat required to convert ice to water + heat required to raise the temperature of water to T degree C

1.1 * 10^5 = 14472 + 60120 + 0.18 * 4190 * (T - 0)

final temperature T = 46.947 degree C