Question

how much would the temperature of 275g of water increase if 36.5kj of heat were added

Answer 1

Where: 

Q = heat =36.5 KJ

m = mass = 275 g

Cp = specific heat for water= 4.184 J/g-^oC

$∆T=\frac{Q}{mCp}$

$∆T=\frac{36.5}{275×4.184}×1000=31.72°C$