# How much energy is required to heat 72 g of water by 75^o C?

May 15, 2017

We use the equation $\setminus \Delta H = m C \setminus \Delta T$.

$\setminus \Delta H = 72 \times 4.2 \times 75 = 22 , 680$ $J$ = $22.7$ $k J$

#### Explanation:

The 'specific heat of water' is $4.184$ $J {g}^{-} 1 {K}^{-} 1$, but is often approximated as $4.2$. (Given that we have the data to only two significant digits, $4.2$ is appropriate to use in this case.)

That means it takes $4.2$ $J$ to raise the temperature of one gram of water by one kelvin, which is the same size unit as one degree celsius. Because we're only talking about a change in temperature, we can use kelvin and celsius interchangeably.

In this case, we have a mass, $m$, of $72$ $g$ of water, with a specific heat, $C$, of $4.2$ $J {g}^{-} 1 {K}^{-} 1$ and we raise its temperature by $\setminus \Delta T$, ${75}^{o}$ $C$. The symbol '$\setminus \Delta$' just means 'the change in'.

Over all, then, the change in energy of the water, $\setminus \Delta H$, is $22.7$ $k J$. Since this number is positive, that is an amount of energy we had to put into the water to heat it up.