Learn It: Sample Problems - Specific Heat Capacities

Exam tip:

When solving problems involving specific heat capacities, you only need to know two equations!

1.      Heat = mass  × specific heat capacity × temperature change
             Q = mc Dq

and

2.        Heat lost = Heat gained
              mc Dq  = mc Dq

When reading the question, concentrate on what is losing heat and what is gaining heat - then use the above two equations to solve the problem.

Worked Examples

Question 1

Given that the specific heat capacity of water is 11 times that of copper, calculate the mass of copper at a temperature of 100 °C required to raise the temperature of 200 g of water from 20.0 °C to 24.0 °C, assuming no energy is lost to the surroundings.     

Solution

Heat lost by copper = heat gained by water
          mcuccuDqcu = mwcwDqw
  mcuccu(100 - 24) = 0.200  × 11ccu(24 – 20)
                  76mcu = 8.8
                      mcu = 0.116 kg

Question 2
Three litres of water at 100 °C are added to 15 litres of water at 40 °C. Calculate the temperature of the mixture. Take the mass of 1 litre of water to be 1 kg and the specific heat capacity of water to be 4.2 × 103 J kg -1 K -1 

Solution

Let the temperature of the mixture q
                             Heat lost = heat gained
                            m1c1Dq1 = m2 c2 Dq2
   3 × 4.2 × 103  × (100 – q) = 15  × 4.2  × 103  × (q -40)
Solving for q  gives q  = 50°C      

You try it!


Question 3
1 kg of water at a temperature of 45 °C is mixed with 1.5 kg of alcohol at 20 °C. Find the final temperature of the mixture.

Take the specific heat capacity of water to be 4200 J kg –1 K –1  and the specific heat capacity of alcohol to be 2400 J kg –1 K –1 . Assume no other exchange of heat occurs.

Solution

Let the final temperature of the mixture be q.
        Heat lost by water = heat gained by alcohol
                    mwcwDqw = malcalDqal
   1  × 4200  × (45 – q) = 1.5  × 2400  × (q – 20)
                     (mcDq)w = (mcDq)al
   1  × 4200  × (45 – q) = 1.5  × 2400  × (q – 20)

Solving for q gives q  = 33 °C.   

You try it!


Application of specific heat capacity: storage heaters

Some electrical heaters contain oil which has a high boiling point and a relatively high specific heat capacity. Once the oil has been heated it retains the thermal energy for quite a while, cooling down only slowly.

In storage heaters, a core of high-density thermal blocks (like very hard building blocks), is heated by electric elements during the night. These blocks are preferred to water, despite having a slightly lower specific heat capacity, because they are more dense (and thus more compact) and do not leak. They are heated during the night using low-cost (off-peak) electricity and release the stored thermal energy during the day.

Storage heaters maintain a steady room air temperature over long periods, warming the entire building fabric and reducing the risk of condensation.

Next: Learn It - Experiment 1

 
 
     

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