Heat of fusion and heat of vaporization

Question

How much heat energy is needed to take a 50 gram piece of ice at -10 degrees Celsius and turn it into steam at 120 degrees Celsius?

1. Specific Heat for Ice: 2.04 J/ g times K
2. Specific Heat for Steam: 1.96 J/ g times K

Answer 1

This is a step-wise calculation where each step looks at the heat needed for a particular change. In general, we use q = mC∆T and q = m∆H for no phase change and phase change respectively. (see below).

q = heat; m = mass; C = specific heat; ∆T = change in temperature (doesn't matter if C or K)

Step 1: heat needed to take 50 g of ice at -10 to 0º (no phase change)

q = mC∆T = (50 g)(2.04 J/gº)(10º) = 1020 J

Step 2: heat needed to change 50 g ice at 0º to liquid water at 0º (phase change)

q = m∆H_fusion = (50 g)(334 J/g) = 16,700 J (look up ∆H_fusion, you may find a different value)

Step 3: heat needed to raise temp of 50 g of liquid water to 100º (no phase change)

q = mC∆T = (50 g)(4.184 J/gº)(100º) = 20,920 J

Step 4: heat needed to change 50 g liquid water at 100º to a gas at 100º (phase change)

q = m∆Hvap = (50 g)(2260 J/g) = 113,000 J (look up ∆Hvap, you may find a different value)

Step 5: heat needed to raise temp of 50 g steam from 100º to 120º (no phase change)

q = mC∆T = (50 g)(1.96 J/gº)(20º) = 1960 J

Add them all up:

1020 J + 16,700 J + 20,920 J + 113,000 J + 1960 J = 153,600 J = 153.6 kJ (not including sig. figs.)