Ice weighing 20 kg at a temperature of -20 ° C is immersed in 20 liters of water at a temperature of 70

Ice weighing 20 kg at a temperature of -20 ° C is immersed in 20 liters of water at a temperature of 70 ° C. Will all the ice be straightened out?

ml = 20 kg.

t1 = – 20 ° C.

t2 = 0 ° C.

Vw = 20 l = 0.02 m3.

ρ = 1000 kg / m3.

t3 = 70 ° C.

Cl = 2100 J / kg * ° С.

Cw = 4200 J / kg * ° С.

q = 3.4 * 105 J / kg.

Ql -?

Qw -?

The required amount for melting ice Ql will be the sum: Ql = Q1 + Q2, where Q1 is the amount of heat required to heat ice from temperature t1 to the melting point of ice t2, Q2 is the amount of heat required for melting itself.

Q1 = Сл * ml * (t2 – t1).

Q2 = q * ml.

Ql = 2100 J / kg * ° C * 20 kg * (0 ° C – (- 20 ° C)) + 3.4 * 105 J / kg * 20 kg = 7640000 J.

The amount of heat Qw, which will be released when the water cools down, is expressed by the formula: Qw = Sv * mw * (t3 – t2).

We express the mass of water by the formula: mw = Vw * ρ.

Qw = 4200 J / kg * ° C * 0.02 m3 * 1000 kg / m3 * (70 ° C – 0 ° C) = 5880000 J.

Sl * ml * (t2 – t1) + q * ml “= Qv.

ml “= (Qв – Сл * ml * (t2 – t1)) / q.

ml “= (5880000 J – 2100 J / kg * ° C * 20 kg * (0 ° C – (- 20 ° C)) / 3.4 * 105 J / kg = 14.8 kg.

Answer: only ml “= 14.8 kg of ice will melt.