Question

How much energy is required to change a 40-g ice cube from ice at

−30°C

to steam at 105°C?

Answer 1

40 g x (1 kg / 1000 g) = 0.040 kg

Now then, the main issue here is that there are FIVE heats you need to calculate, then find the sum.

1) The ice warms from -30ºC to its melting point, 0ºC.

Since this is a temperature change, use the formula q = mcΔT

q = (0.040 kg)(2090 J/kgºC)(30ºC) = 2508 J

2) The ice melts while the temperature holds steady at 0ºC.

Since this is a phase change, use the formula q = mL. Since this phase change is melting, use the heat of fusion for L, which is 3.34x10^5 J/kg.

q = (0.040 kg)(3.34x10^5 J/kg) = 13360 J

3) The meltwater warms from 0ºC to 100ºC.

Another temperature change, but this time it's water (rather than ice), so you'll need to use the specific heat of water instead.

q = (0.040 kg)(4186 J/kgºC)(100ºC) = 16744 J

4) The water boils while the temperature holds steady at 100ºC.

Another phase change. Use the heat of vaporization and the formula q = mL

q = (0.040 kg)(2.26x10^6 J/kg) = 90400 J

5) The steam warms from 100ºC to 105ºC.

One last temperature change. The specific heat of steam is about 2010 J/kgºC (depending on the temperature and pressure).

q = (0.040 kg)(2010 J/kgºC)(5ºC) = 402 J

Add the five heats together and you get 123414 J.