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

1 Answer
May 15, 2017

Answer:

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.