Question

1. How much energy is required to change a 41 g ice cube from ice at −14◦C to steam at 115◦C? The specific heat of ice is 2090 J/kg · ◦ C, the specific heat of water is 4186 J/kg · ◦ C, the specific heat of stream is 2010 J/kg · ◦ C, the heat of fusion is 3.33 × 105 J/kg, and the heat of vaporization is 2.26 × 106 J/kg. Answer in units of J.

Answer 1

When there is no phase change, heat required = mcT, where m is mass, c is specific heat, T is change in temperature.

When there is phase change, heat required = mL, where m is mass, L is latent heat of the phase change.

Here,m=41 g = 0.041 kg.

For ice, c=2090 J/kg. So, to increase temperature of ice from -14 deg. celsius to 0 deg. celsius, heat required = 0.041*2090*14 = 1199.66 J.

Now, to convert ice  into water at 0 deg. celsius, heat required = mL, where L is latent heat of fusion = 333000 J/kg. So, heat required = 0.041*333000 = 13653 J.

For water, c=4186 J/kg. So, to increase temperature of water from 0 deg. celsius to 100 deg. celsius, heat required = 0.041*4186*100 = 17162.6 J.

Now, to convert water into steam at 100 deg. celsius, heat required = mL, where L is latent heat of vaporization = 2260000 J/kg. So, heat required = 0.041*2260000 = 92660 J.

For steam, c=2010 J/kg. So, to increase temperature of steam from 100 deg. celsius to 115 deg. celsius, heat required = 0.041*2010*15 = 1236.15 J.

So, total heat required = 1199.66+13653+17162.6+92660+1236.15 = 125911.41 J