Multiple changes

Specific heat capacity relates only to the energy required for a change in temperature. Specific latent heat relates only to the energy required for a change in state. If a change in internal energy 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)?

  • 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 Q = mc\Delta \theta \

 \ \Delta Q = 1 \times 4,200 \times 25 \

 \ \Delta Q = 105,000 \ J \

  • 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:

\ Q \ = \ ml \

\ Q \ = \ 1 \times 2,260,000 \ J \

\ Q \ = \ 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 1,859 J/kg°C, the final temperature of the steam would be:

 \ \Delta Q = 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.

  • Raising the temperature of the water to 100°C. So the amount of energy needed in this case would be:

 \ \Delta Q = mc\Delta \theta \

 \ \Delta Q = 0.5 \times 4,200 \times 20 \

 \ \Delta Q = 42,000 \ J \

  • Turning the water into steam. So the amount of energy needed in this case would be:

\ Q \ = \ ml \

\ Q \ = \ 0.5 \times 2,260,000 \ J \

\ Q \ = \ 1,130,000 \ J \

  • Raising the temperature of the steam from 100°C. So energy needed in this case would be:

 \ \Delta Q = mc\Delta \theta \

 \ \Delta Q = 0.5 \times 1,859 \times 10 \

 \ \Delta Q = 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