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

1 Answer
May 15, 2017

We use the equation \DeltaH=mC\DeltaT.

\DeltaH=72 xx 4.2 xx 75 = 22,680 J = 22.7 kJ

Explanation:

The 'specific heat of water' is 4.184 Jg^-1K^-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 Jg^-1K^-1 and we raise its temperature by \DeltaT, 75^o C. The symbol '\Delta' just means 'the change in'.

Over all, then, the change in energy of the water, \DeltaH, is 22.7 kJ. Since this number is positive, that is an amount of energy we had to put into the water to heat it up.