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Specific heat /heat capacity : Examples

Here are some typical problems and exercises on the topic specific heat .

1) How much heat is needed to raise the temperature of a block of copper ( weighing 0.5 kg) from 0°C to 100° C  ? (for copper, c = 386 J / kg oC)

2) How much heat is needed to raise the temperature of 0.5 kg of water from 0°C to 100° C? (for water, c = 4186 J / kg oC)

3)What would be the final temperature of a mixture of 100 g of water at 90°C and 600 g of water at 20°C ?

4)What would be the final temperature if a 2 kg piece of lead at 200°C is inserted in a container with 10 kg of water at 50°C ? (for lead, c = 128 J / kg oC)

Examples involving melting ice are a bit more tricky and can be found here: exercises involving also latent heat>>


1) Applying the formula:

Q = 386 * 0.5 * 100 = 19300 J or 19.3 kJ

Comments: It is important to observe the SI units. The mass is in kg and the heat energy in J. Normally the temperature is converted into K, but because we are taking the difference (or the variation), it doesn't matter what units are used (if kelvin or celsius). However, if the difference of temperatures can be plugged in the formula in the scales kelvin or celsius (no need to convert celsius into kelvin). If the difference is in degrees farenheit a conversion (to celsius or kelvin) is needed before using the formula.


2) Q = 4186 * 0.5 * 100 = 209300 J or 209.3 kJ

Comments: Note that this is more than 1o times the nergy needed in the case of copper!

3) This example is also typical and it requires some algebraical skills.

We know that heat flows from the hotter body to the cooler body. Hence, the water at the higher temperature will "loose" heat and the water at the lower initial temperature will "gain" heat. The correct way of describing this situation is by saying that heat will be transferred from the hotter to the cooler water.

We also know that, by conservation of energy, the amount of heat lost will be the same that is gained.

So, let's call the final temperature of the mixture Tf.

The amount of heat that will be transferred from the hotter water is:

4186 * 0.1 * (90 - Tf)

The amount of heat that will be transferred to the cooler water is:

4186 * 0.6 * ( Tf -20)

Because these two quantities must be equal, we have an equation:

4186 * 0.1 * (90 - Tf) = 4186*0.6 * (Tf - 20)

We need to find Tf:

418.6 * (90-Tf) = 2511.6 * (Tf-20)

Getting rid of the brackets:

37674 - 418.6 Tf = 2511.6 Tf - 50232

-2930.2 Tf = -87906

Tf = 30° C

4) Similar to 3) above,

in this case:

amount of heat transferred from the lead:


amount of heat transferred to the water:


Equating the two heats:

128*2*(200-Tf) = 4186*10*(Tf-50)


51200-256 Tf = 41860 Tf - 2093000

42116 Tf = 2144200

Tf = 50.9°


exercises involving also latent heat>>