Question

How much heat must be removed by a refrigerator from $4\text{kg}$ of water at ${70}^{\circ }\text{C}$ to convert it to ice cube at $-{10}^{\circ }\text{C}$ ?
[Take specific heat of water $C_w=4200{\text{J/kg/}}^{\circ }\text{C}$, Latent heat of fusion of ice$L_f=334000\text{J/kg}$, Specific heat of ice $C_i=2100\text{J/kg K}$]

• A. $3521\text{kJ}$
• B. $2596\text{kJ}$
• C. $2352\text{kJ}$
• D. $4200\text{kJ}$

Answer 1

The correct option is B $2596\text{kJ}$

Energy removed when temperature of water changes from ${70}^{\circ }\text{C}$ to $0^{\circ }\text{C}$
$Q_1=m{c}_w\Delta T$
$=4\times 4200\times (70-0)$
$=4\times 4200\times 70=1176000\text{J}$
Energy removed to freeze $4\text{kg}$ of water.
$Q_2=m{L}_w=4\times 334000$
$Q_2=1336000\text{J}$
Energy to be removed to reduce the temperature of ice from $0^{\circ }\text{C}$ to $-{10}^{\circ }\text{C}$
$Q_3=m{c}_i\Delta T$
$=4\times 2100\times (0-(-10)$
$=4\times 2100\times 10=84000\text{J}$
Therefore, total heat removed
$=Q_1+Q_2+Q_3=1176000+1336000+84000$
$=2596000\text{J}=2596\text{kJ}$
Hence, option (b) is correct.