Question

How much ice at -10C must be added to 0.50 Kg of water at 20 C if it all to melt leaving only water at 0 C?

Answer 1

Let's construct heat ballance equation:
C_ice*m*(T₁-T₀) + C_fnice * m = C_water*m_water*(T₂-T₀)

where C_ice is a specific heat of ice, 2.108 J/gK , C_water is a specific heat of water of 4.187 J/gK, C_fnice - the heat of ice fusion 333.55 J/g, T₀ = 0C (melting point of ice), T₁ = -10C, T₂ = 20C, m_water = 0.5kg=500g.
21.08m + 333.55 m = 41870

m = 118.1 g = 0.118 kg of ice.