Question

3.00 L of water is heated to 96.00°C by the addition of 52.70kJ of heat. What was the initial heat of the temperature?

Answer 1

From the heat absorbed equation

We have

$q=mc(T_2-T_1)$

$m=2000\ g\\c=4.184\ J\ g^{-1}\ \degree C^{-1}$

$T_2=369\ K\\q=52700\ \ J$

$T_2-T_1=\frac{q}{mc}=\frac{52700}{2000\times 4.184}=6.297\ K$

$T_1=T_2-6.297=369-6.297=362.7\ K$

Initial temprature$\ =362.7\ K$