What is the final temperature of 400 g of water at 20°C after it absorbs 226 kJ of heat?

1 Answer
May 26, 2016

Answer:

I found: #T_f=155^@C#

Explanation:

I would use the relationship:
#Q=mcDeltaT#
where:
#Q# is the heat;
#m# is the mass;
#c# is the specific heat;
#DeltaT=T_f-T_i# is the difference in temperature.
From literature we have:
#c_("water")=4.18J/(g^@C)#
So we have:
#226xx10^3=400*4.18(T_f-20)#
#T_f=155^@C#