Answer:[tex]19 ^{\circ}[/tex]
Explanation:
Given
mass of water [tex](m_w) 3.95 kg[/tex]
Temperature of water is [tex]20^{\circ} C[/tex]
mass of ice[tex](m_i) =45 gm[/tex]
Temperature of ice [tex]=-10.2 ^{\circ} C[/tex]
specific heat of water [tex]4.18 kJ/kg-K[/tex]
specific heat of ice [tex]2.22 kJ/kg-K[/tex]
Latent heat of fusion(L) [tex]3.4\times 10^5 J/kg[/tex]
Heat absorb by water when ice completely converts to water
[tex]Q=mc_{ice}\delta T+mL[/tex]
[tex]Q=0.045\times 2.22\times (0+10.2)+0.045\times 3.4\times 100[/tex]
Q=1.01898+15.3=16.3189 KJ
thus temperature of water will be
[tex]3.95\times 4.18\times (20-T)=16.3189[/tex]
[tex]T=19.011\approx 19^{\circ} C[/tex]
