Determine how much heat is needed to convert 200g of ice taken at a temperature of 0 degrees into steam at 100 degrees

Determine how much heat is needed to convert 200g of ice taken at a temperature of 0 degrees into steam at 100 degrees. Specific heat of melting of ice 340 kJ / kg, specific heat of water 4.2 kJ / kg * С, specific heat of vaporization of water 2300 kJ / kg

Q = Q1 + Q2 + Q3.

Q1 = λ * m, where λ – beats. heat of melting of ice (λ = 340 * 10³ J / kg), m – ice mass (m = 200 g = 0.2 kg).

Q2 = C * m * (tк – tн), where C – beats. heat capacity of water (C = 4200 J / (K * kg)), tc – end. pace. (tк = 100 ºС), tн – early. pace. (tн = 0 ºС).

Q3 = L * m, where L – beats. heat of vaporization of water (L = 2.3 * 10 ^ 6 J / kg).

Q = Q1 + Q2 + Q3 = λ * m + C * m * (tк – tн) + L * m = 340 * 10³ * 0.2 + 4200 * 0.2 * (100 – 0) + 2.3 * 10 ^ 6 * 0.2 = 612000 J = 612 kJ.