Into a pan with ice, the mass of which is equal to ml = 200 g, and the temperature tx = 0 ° С, water was poured

Into a pan with ice, the mass of which is equal to ml = 200 g, and the temperature tx = 0 ° С, water was poured, the mass of which is equal to m2, and the temperature / 2 = 50 ° С. After the ice had melted, the temperature in the vessel was set at t3 = 7 ° C. The mass of the poured water is equal to. The specific heat capacity of water c is 4.2 kJ / kg-K, the specific heat of melting of ice is 330 kJ / kg, and the heat capacity of the vessel is neglected.

ml = 200 g = 0.2 kg.
tl = 0 ° C.
tv1 = 50 ° C.
t2 = 7 ° C.
Cw = 4.2 kJ / kg * ° C = 4200 J / kg * ° C.
λ = 330 kJ / kg = 330,000 J / kg.
mv -?
For ice melting, the amount of heat Q1 = λ * ml is required.
To heat the water obtained by melting ice, you need the amount of heat Q2 = Cw * ml * (t2 – tl).
When the water that was poured was cooled, the amount of heat Q3 = Cw * mw * (t1 – t2) was released.
Let’s write down the heat balance equation: Q3 = Q1 + Q2.
Cw * mw * (t1 – t2) = λ * ml + Cw * ml * (t2 – tl).
mv = (λ * ml + Cw * ml * (t2 – tl)) / Cw * (t1 – t2).
mw = (330,000 J / kg * 0.2 kg + 4200 J / kg * ° C * 0.2 kg * (7 ° C – 0 ° C)) / 4200 J / kg * (50 ° C – 7 ° C ) = 0.398 kg.
Answer: mw = 0.398 kg of water was poured into the ice.