Multiple changes
specific heat capacityThe amount of energy needed to raise the temperature of 1 kg of substance by 1°C. relates only to the energy required for a change in temperature. specific latent heatThe amount of energy needed to melt or vaporise 1 kg at its melting or boiling point. relates only to the energy required for a change in state. If a change in internal energyThe total kinetic energy and potential energy of the particles in an object. of a material will cause it to change temperature and change state, both equations can be used.
Example
What happens when 1 kilogram (kg) of water at 75 degrees Celsius (°C) is heated with 2.5 megajoules (MJ) (2,500,000 J)?
1. Some of the energy is used to raise the temperature of the water to 100°C, so the energy needed to raise 1 kg of water by 25°C is:
\(\Delta QE = mc \Delta \theta\)
\(\Delta QE = 1 \times 4,200 \times 25\)
\(\Delta QE = 105,000~J\)
2. Some of the remaining 2,395,000 J is then used to turn the boiling water into steam, so the energy needed to change 1 kg of water at 100°C into steam at the same temperature is:
\(\Delta QE = mL\)
\(\Delta QE = 1 \times 2,260,000~J\)
\(\Delta QE = 2,260,000~J\)
The final amount of energy 2,500,000 - 2,260,000 - 105,000 = 135,000 J, is used to raise the temperature of the steam, and as steam has a specific heat capacity of 1859 J/kg°C, the final temperature of the steam would be:
\(E = mc \Delta \theta\)
\(135,000 = 1 \times 1,859 \times \Delta \theta\)
\(\Delta \theta = \frac{135,000}{1 \times 1,859}\)
\(\Delta \theta = 72.6~ {\textdegree}C\)
The steam started at 100°C and heats up by 72.6°C so is now 172.6°C.
Question
If 0.5 kg of water at 80°C is changed into steam at 110°C, how will the energy be used?
Energy will go into three places.
1. Raising the temperature of the water to 100°C. So the amount of energy needed in this case would be:
\(E = mc \Delta \theta\)
\(E = 0.5 \times 4,200 \times 20\)
\(E = 42,000~J\)
2. Turning the water into steam. So the amount of energy needed in this case would be:
\(E = mL\)
\(E = 0.5 \times 2,260,000~J\)
\(E = 1,130,000~J\)
Raising the temperature of the steam from 100°C. So energy needed in this case would be:
\(E = mc \Delta \theta\)
\(E = 0.5 \times 1,859 \times 10\)
\(E = 9,295~J\)
So the total amount of energy needed to change 0.5 kg of water at 80°C into steam at 110°C would be:
Total amount of energy = 42,000 + 1,130,000 + 9,295 = 1,181,295 J