A thin-walled vessel contains a mixture of ice and water at a temperature of 0 ° C. The mass of ice is 350 g

A thin-walled vessel contains a mixture of ice and water at a temperature of 0 ° C. The mass of ice is 350 g, and the mass of water is 550 g. The vessel is heated on a 1.5 kW burner. How long does it take to bring the contents of the vessel to a boil? Heat loss and specific heat capacity of the vessel, as well as water evaporation can be neglected.

To find the heat required to bring the mixture to a boil, we will use the equality: N * t = Q (heat) = λ * ml + Sv * (ml + mv) * (tboil – t), whence t = (λ * ml + Sv * (ml + mv) * (tboil – t)) / N.

Constants and variables: λ – specific heat of ice melting (λ = 3.4 * 105 J / kg); ml is the mass of ice (ml = 350 g = 0.35 kg); Sv – specific heat capacity of water (Sv = 4200 J / (kg * K)); mw is the mass of water (mw = 550 g = 0.55 kg); tboil – boiling point (tboil = 100 ºС); t – temperature of ice and water before heating (t = 0 ºС); N – burner power (N = 1.5 kW = 1500 W).

Calculation: t = (λ * ml + Sv * (ml + mv) * (tboil – t)) / N = (3.4 * 105 * 0.35 + 4200 * (0.35 + 0.55) * ( 100 – 0)) / 1500 = 331.3 s ≈ 5.52 min.

Answer: It will take 5.52 minutes of time.