Question

How much heat is required to change a 30.2 g ice cube from ice at -13.8°C to water at 23°C? (if necessary, use c_ice=2090 J/kg°C and c_steam= 2010 J/kg°C)

How much heat is required to change a 30.2 g ice cube from ice at -13.8°C to steam at 120°C?

Answer 1

m = 0.0302 gm

1) The ice warms from -13.8ºC to its melting point, 0ºC.

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

q = (0.0302 kg)(2090 J/kgºC)(13.8ºC) = 871 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 = (0.0302 kg)(3.34x10^5 J/kg) = 10086.8 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 = (0.0302 kg)(4186 J/kgºC)(100ºC) = 12641.72 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 = (0.0302 kg)(2.26x10^6 J/kg) = 68252 J

5) The steam warms from 100ºC to 120ºC.

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

q = (0.0302 kg)(2010 J/kgºC)(20ºC) = 1214 J

Add the five heats together and you get 93065.52 J

conversion of ice to water is quite easy for using this solution.