#1.2 xx 10^5# #J# of thermal energy are added to a sample of water and its temperature changes from #229# #K# to #243# #K#. What is the mass of the water?

1 Answer
David G.
Mar 23, 2016

#Delta H = mCDeltaT#

Rearranging:

#m = (Delta H)/(CDeltaT)=(1.2xx10^5)/(4181xx(243-229))=2.1# #kg#

Explanation:

We use the equation #Delta H = mCDeltaT# where #m# is the mass #(kg)#, #T# is the temperature #(K)#, #H# is the thermal energy #(J)# and #C# is the specific heat #(Jkg^-1K^-1)#. #Delta# just means 'the change in'.

We know that for water, #C=4181# #Jkg^-1K^-1#. (Sometimes quoted as #4.2# #Jg^-1K^-1#, but it's better to work in SI units.)

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