Answer:209.98 kJ
Explanation:
mass of water [tex]m=456 gm[/tex]
Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]
Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]
specific heat of ice [tex]c=2090 J/kg-K[/tex]
Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]
specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]
Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]
[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]
Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]
[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]
heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]
[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]
Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]
[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]
