# 37. Specific heat capacity – calculation -

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## Specific heat capacity - calculation

If the specific heat capacity of water is 4200 J/kg°C and a kettle boils 500g of water from 20°C to 100°C, how much heat energy does the kettle transfer to the water?

### 1. First we need to LIST the data we know and the data we want to find.

Note that it is best to use the letter symbols that represent each quantity, because that will help us to identify a equation that we can use.

c = 4200 J/kg°C
m = 500g
θ = +80°C
E = ? J

Before we continue, we need to check that the units AGREE. We can see that we have grams for mass, but kilograms for specific heat capacity. So, we need to convert grams to kilograms (which is a standard unit in physics) remembering that there are 1000g in 1kg:

c = 4200 J/kg°C
m = 500g = 0.5kg
θ = +80°C
E = ? J

### 2. Now it's time to state the EQUATION that links these data together

(We're still using letter symbols)

E = m × c × θ

### 3. SOLVE the equation by putting the numbers in:

E = 0.5 × 4200 × 80 = 168000

### 4. Now STATE the answer with the correct unit:

Energy required = 168000 J

And that's it! Remember to use List - Equation - Solve - State for all your calculations to gain valuable marks in your exam!

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