Question

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090 J/(kg.K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg .

Answer 1

Answer:209.98 kJ

Explanation:

mass of water $m=456 gm$

Initial Temperature of Water $T_i=25^{\circ}C$

Final Temperature of water $T_f=-10^{\circ}C$

specific heat of ice $c=2090 J/kg-K$

Latent heat $L=33.5\times 10^4 J/kg$

specific heat of water $c_{water}=4.184 KJ/kg-K$

Heat require to convert water at $T=25^{\circ}C$ to $T=0^{\circ}C$

$Q_1=0.456\times 4.184\times (25-0)=47.69 kJ$

Heat require to convert water at $T=0^{\circ}$ to ice at $T=0^{\circ}$

$Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ$

heat require to convert ice at $T=0^{\circ} C\ to\ T=-10^{\circ} C$

$Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ$

Total heat $Q=Q_1+Q_2+Q_3$

$Q=47.69+152.76+9.53=209.98 kJ$