Question

Calculate the heat absorbed by the 100mL sample of water. The specific heat capacity of water is 4.18J/g℃.

Answer 1

Specific heat capacity is the amount of heat required per gram of water to raise its temperature by $1\degree C$ .

$C=4.18J/g \ \degree C$

Mass of water taken is $100 \ ml$ and water density is $1g/ml$ .

So, $m=100\ g$ .

If the water temperature rises from its initial temperature $T_1$ to final temperature $T_2$ , then change in temperature $\triangle T$ $=(T_2-T_1)$ .

Heat absorbed is given by $Q=mC\triangle T=mC(T_2-T_1)==100\times 4.18\triangle T\ J=418\triangle T \ J$

If temperature change is $\triangle T=50\degree C,$ then

$Q=418\times 50\ J=20900J=20.9\ KJ$