Multiple thermal changes
specific heat capacityThe amount of energy needed to raise the temperature of 1 kg of substance by 1掳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 (掳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掳C, so the energy needed to raise 1 kg of water by 25掳C is:
\(\Delta Q E= mc \Delta \theta\)
\(\Delta Q E= 1 \times 4,200 \times 25\)
\(\Delta Q E= 105,000~J\)
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掳C into steam at the same temperature is:
\(\Delta Q E = mL\)
\(\Delta Q E = 1 \times 2,260,000~J\)
\(\Delta Q E = 2,260,000~J\)
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掳C, the final temperature of the steam would be:
\(\Delta Q E = mc \Delta \theta\)
\(135,000 = 1 \times 1,859 \times \Delta \theta\)
\(\Delta \theta = \frac{135,000}{1 \times 1,859}\)
\(\Delta \theta = 72.6~ {\textdegree}C\)
The steam started at 100掳C and heats up by 72.6掳C so is now 172.6掳C.
Question
If 0.5 kg of water at 80掳C is changed into steam at 110掳C, how will the energy be used?
Energy will go into three places.
1. Raising the temperature of the water to 100掳C. So the amount of energy needed in this case would be:
\(\Delta QE = mc \Delta \theta\)
\(\Delta QE = 0.5 \times 4,200 \times 20\)
\(\Delta QE = 42,000~J\)
2. Turning the water into steam. So the amount of energy needed in this case would be:
\(QE = mL\)
\(QE = 0.5 \times 2,260,000~J\)
\(QE = 1,130,000~J\)
3. Raising the temperature of the steam from 100掳C. So the amount of energy needed in this case would be:
\(\Delta QE= mc \Delta \theta\)
\(\Delta QE = 0.5 \times 1,859 \times 10\)
\(\Delta QE = 9,295~J\)
So the total amount of energy needed to change 0.5 kg of water at 80掳C into steam at 110掳C would be:
Total amount of energy = 42,000 + 1,130,000 + 9,295 = 1,181,295 J