Question

How much energy is required to change a 23 g ice cube from ice at − 12 ◦ C to steam at 101 ◦ C? The specific heat of ice is 2090 J / kg · ◦ C and of water 4186 J / kg · ◦ C. The latent heat of fusion of water is 3 . 33 × 10^5 J / kg, its latent heat of vaporization is 2 . 26 × 10^6 J / kg, and the specific heat of steam is 2010 J / kg · ◦ C. Answer in units of MJ.

Answer 1

We can break this up into 5 stages.

Stage 1 = Heating the ice to 0 degrees.
Stage 2 = Melting the ice.
Stage 3 = Heating the water to 100 degrees.
Stage 4 = Turning the water into steam.
Stage 5 = Heating the steam to 101 degrees.

The energy required is as follows:
Stage 1 : For ice at 12 to ice at 0 = (0.023 kg)(2090 J/kgC)(12 C) = 576.84 J
Stage 2 : For ice at 0 to water at 0 = (0.023 kg)(3.33 x 10^5 J/kg) = 7659 J
Stage 3 : For water at 0 to water at 100 = (0.023 kg)(4186 JkgC)(100C) = 9627.8 J
Stage 4 : For water at 100 to steam at 100 = (0.023 kg)(2.26 x 10^6 J/kg) = 51980 J
Stage 5 : For steam at 100 to steam at 101 = (0.023 kg)(2010 J/kgC)(1C) = 46.23 J

Adding all these up gives a total energy of 69889.87 J = 0.0698 MJ