Answer: [tex]7.95^{\circ}C[/tex]
Explanation:
Given
Mass of water is [tex]m_1=1\ kg[/tex]
mass of ice is [tex]m_2=0.1\ kg[/tex]
Latent heat of fusion [tex]L=3.33\times 10^5\ kJ/kg[/tex]
The heat capacity of water is [tex]c_{w}=4186\ J/kg^{\circ}C[/tex]
Suppose water is at [tex]T^{\circ} C[/tex] and it reaches to [tex]0^{\circ}C[/tex] to melt the ice
the heat released by water must be equivalent to heat absorbed by the ice
[tex]\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C[/tex]