Question

2 kg of water at 293 K is converted completely into ice at 273 K. Calculate the heat liberated.

• A. 840 kJ
• B. 8.4 kJ
• C. 8400 kJ
• D. 84 J

Answer 1

The correct option is A 840 kJ

Given that, m = 2 kg

$T_1=293K,C=4.2\times {10}^{3}Jk{g}^{-1}{K}^{-1}$

$L_f=336\times {10}^{3}J/kg$

First water at 293 K is cooled down to 273 K and then freezes.
Heat liberated during cooling from 293 K is given by

$Q_1=mc\Delta T$

$Q_1=2\times 4.2\times {10}^3\times (293-273)$

$Q_1=168\times {10}^3\text{Joules}$

Heat liberated during freezing at 273 K is

$Q_2=m{L}_f$

$=2\times 336\times {10}^3\text{Joules}$

$=672\times {10}^3\text{Joules}$

Total heat liberated $Q_1+Q_2$

$=(168+672)\times {10}^3\text{Joules}$

$=840\times {10}^{3}J=840kJ$