What would be the final temperature if #3.31 * 10^3# joules of heat were added to 18.5 grams of water at 22.0°C?

1 Answer
Jan 30, 2016

Answer:

I found: #T_f=64.6^@C#

Explanation:

We can try using the general relationship between heat #Q# (lost or gained) and change in temperature #DeltaT# of a mass #m# of a substance of specific heat #c# (in our case water: #c_("water")=4.2J/(g^@C)#):

#Q=mcDeltaT#

with our data:
#3.31xx10^3=18.5*4.2*(T_f-22)#
rearranging we get:
#T_f=64.6^@C#