Answer:
Explanation:
Given
mass of water [tex]m_1=0.27\ kg[/tex]
Temperature of water [tex]T_{wi}=82.5^{\circ}C[/tex]
Initial Temperature of ice[tex]=-22.3^{\circ}C[/tex]
Final temperature of system [tex]T=34^{\circ}C[/tex]
specific heat of water [tex]c=4.18\ kJ/kg-K[/tex]
specific heat of ice [tex]c_i=2.108\ kJ/kg-K[/tex]
Latent heat of ice [tex]L=336\ kJ/kg[/tex]
Heat loss by Water is equal to heat gained by ice
Heat loss by water [tex]Q_1=m_w\times c\times \Delta T[/tex]
[tex]Q_1=0.27\times 4.18\times (82.5-34)=54.7371\ kJ[/tex]
Heat gained by ice [tex]Q_1=x\times c_i(0-(-22.3))+x\times L+x\times c\times (T-0)[/tex]
[tex]Q_2=x\times 2.108\times (22.3)+x\times 336+x\times 4.18\times 34[/tex]
[tex]Q_2=525.1284x\ kJ[/tex]
[tex]Q_1=Q_2[/tex]
[tex]x=\frac{54.73}{525.1284}[/tex]
[tex]x=0.104\ kg[/tex]
