# A piece of ice m = 2.0 kg at a temperature of t = -20 (degrees Celsius) was heated, giving it the amount

**A piece of ice m = 2.0 kg at a temperature of t = -20 (degrees Celsius) was heated, giving it the amount of heat Q = 1.26 MJ. The temperature of the substance after heating is?**

m = 2 kg.

Cl = 2100 J / kg * ° C.

q = 3.3 * 10 ^ 5 J / kg.

Cw = 4200 J / kg * ° C.

t1 = -20 ° C.

t2 = 0 ° C.

Q = 1.26 MJ = 1.26 * 10 ^ 6 J.

t3 -?

Let’s find the amount of heat Q1, which is necessary to heat the ice to the melting temperature, which is t2 = 0 ° C according to the formula: Q1 = Сл * m * (t2 – t1).

Q1 = 2100 J / kg * ° C * 2 kg * (0 ° C – (-20 ° C)) = 84000 J.

Let us find the amount of heat Q2, which is required to melt ice at the melting temperature, according to the formula: Q2 = q * m.

Q2 = 3.3 * 10 ^ 5 J / kg * 2 kg = 660,000 J.

Let’s find the amount of heat Q3, which goes to heating the resulting water: Q3 = Q – Q1 – Q2.

Q3 = 1,260,000 J – 84,000 J – 660,000 J = 516,000 J.

The amount of heat Q3, which is necessary for heating water, is expressed by the formula: Q3 = Sv * m * (t3 – t2).

t3 – t2 = Q3 / Sv * m.

t3 = t2 + Q3 / Sv * m.

t3 = 0 ° C + 516000 J / 4200 J / kg * ° C * 2 kg = 61.4 ° C.

Answer: after heating, the water temperature was t3 = 61.4 ° C.