Question

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

Answer 1

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`