What is the initial temperature of water with a mass of 800 g of ice, if in order to raise its temperature to 0 degrees

What is the initial temperature of water with a mass of 800 g of ice, if in order to raise its temperature to 0 degrees it was necessary to increase its internal energy by 33.6 kJ?

Problem data: m (mass of heated ice) = 800 g (in SI m = 0.8 kg); t (final temperature of ice heating) = 0 ºС; ΔQ (increase in the internal energy of ice) = 33.6 kJ (33.6 * 10 ^ 3 J).

Reference values: Cl (specific heat of ice) = 2100 J / (kg * K).

Let us express the initial temperature of the taken ice from the formula: ΔQ = Сl * m * (t – t0), whence t0 = t – ΔQ / (Сl * m).

Let’s calculate: t0 = 0 – 33.6 * 10 ^ 3 / (2100 * 0.8) = -20 ºС.

Answer: Before the transfer of heat, the ice had a temperature of -20 ºС.