Question

An ice cube of volume 15 cm³ is initially at the temperature -20C . How much heat is required to convert this ice cube into steam?

Answer 1

1. How much heat is needed to raise the temperature of the ice from -20 to 0.
2. How much heat is required to turn the ice into water.
3. The heat needed to raise the temperature of the water from 0 to 100
4. The amount of heat needed to turn the water into steam.
The problem isn't hard, just a bit tedious so here goes..

1. We will use the equation Q=mc(dT) c is a constant that we can look up, we will have to know the mass of the ice cube and convert our temperature into Kelvin.
Knowing to volume of the ice and its density (Just need to look it up) We can use the equation of density.
p=m/v ---> m=pv=(.91 g/cm^3)(15 cm^3)=13.8 g
Now we need to convert Celsius into Kelvin, just add 273 since we are going from Celsius to Kelvin
-22.2 C=253K and 0 C=273 K so then our dt=273K-253=20kNow plug in our numbers
Q=(13.8g)(2.05J/gK)(20K)=563.83 J That's step one.

2. This step we will need to know the latent heat of fusion because we are going from ice to water. This value should be a table and we will use the equation Q=mL mass we know already and L we will look up.
Q=(13..8x10^-3 kg)(334 kJ/kg)= 4.6 kJ

3. This step is the same as step one expect we have the change our value for c because we are dealing with water and not ice.
Q=(13.8g)(4.2 J/gK)(100 K) Since 373-273=100K
Q=5796J

4. Finally, converting all that water into steam..this time we have to use the heat of vaporization..
Q=ml=(13.8x10^-3 kg)(2260 kJ/kg)=31.8 kJ
Now, just add up the values you got from steps 1-4, note I've already converted kJ to J
Q(Total)=563.83 J +4600 J + 5796J + 31800 J=42759.83 J = 42.7KJ