Question

How much heat energy is released when $5kg$ of water at ${20}^{\circ }C$ is brought to its freezing point?
$(\text{Specific heat of water}=4.2kJk{g}^{-1}{c}^{-1})$

• A. $240kJ$
• B. $1680kJ$
• C. $420kJ$
• D. $168kJ$

Answer 1

The correct option is C $420kJ$

Given:
Mass of water, $m=5kg$
Specific heat of water, $c=4.2kJk{g}^{-1}{}^{\circ }{C}^{-1}$
Fall in temperature of water $\Delta \theta =0-20=-{20}^{\circ }C$

Heat released is given by:
$Q=mc\Delta \theta$
$Q=5\times 4.2\times {10}^3\times (-20)J$
Negetive sign indicates that energy is released
$Q=420\times {10}^{3}J=420kJ$