500 g of ice at a temperature of -10 ° was turned into water with a temperature of 50 °.

500 g of ice at a temperature of -10 ° was turned into water with a temperature of 50 °. How much heat was required for this? (Specific heat of melting of ice 34 * 10 ^ 4 J / kg, specific heat of ice 2100 J / (kg * 0C), specific heat of water 4200 J / (kg * 0C))

Q = Q1 + Q2 + Q3.
Q1 = C1 * m * (tк1 – tн1), С1 – beats. warm ice (C1 = 2100 J / (kg * K)), m – mass (m = 0.5 kg), tc – end. pace. (tк = 0 ºС), tн – early. pace. (tн = -10 ºС).
Q2 = λ * m, λ – beats. melting heat ice (λ = 34 * 10 ^ 4 J / kg).
Q3 = C2 * m * (tк2 – tн2), C2 – beats. warm water (C2 = 4200 J / (K * kg)), tк2 – end. pace. (tк2 = 50 ºС), tн2 – early. pace. (tн2 = 0 ºС).
Q = 2100 * 0.5 * 10 + 34 * 10 ^ 4 * 0.5 + 4200 * 0.5 * 50 = 10500 + 170,000 + 105000 = 285500 J = 285.5 kJ.