Answer to Question #159210 in Chemistry for Mya

Question

How much energy is required to change 25 g of ice at -20 degrees C into water at 20 degrees C?

Answer 1

Solution:

This problem requires several steps:

Step 1: raise 25 g ice from -20ºC to 0ºC

  q = mC∆T = (25 g) × (2.108 J g^-1 deg^-1) × (20 deg) = 1054 J

 

Step 2: melt 25 g of ice at 0ºC

  q = m∆H_f = (25 g) × (334 J/g) = 8350 J

 

Step 3: heat 25 g of water from 0º to 20º

  q = mC∆T = (25 g) × (4.187 J g^-1 deg^-1) × (20 deg) = 2093.5 J

 

Add up all the heat required for all these changes is

1054 J + 8350 J + 2093.5 J = 11497.5 J = 11.4975 kJ = 11.5 kJ

So, changing 25 g of ice at -20ºC into 25 g into water at 20ºC would take approximately 11.5 kJ of heat energy.

Answer: 11.5 kJ of heat energy is required.