In a bucket containing 5 liters of water at a temperature of 20ºC, 1.5 kg of ice at a temperature of 0ºC was thrown

In a bucket containing 5 liters of water at a temperature of 20ºC, 1.5 kg of ice at a temperature of 0ºC was thrown. Find the temperature of the water in the bucket when thermal equilibrium is established.

V = 5 l = 0.005 m3.

ρ = 1000 kg / m3.

t1 = 20 ºC.

ml = 1.5 kg.

t2 = 0 ºC.

C = 4200 J / kg * ºC.

q = 3.4 * 10 ^ 5 J / kg.

t -?

To melt ice, you need the amount of heat Ql, which is determined by the formula: Ql = q * ml.

Ql = 3.4 * 10 ^ 5 J / kg * 1.5 kg = 510,000 J.

When the water cools down from t1 to t2, the amount of heat will be released, which is determined by the formula: Qw = C * m * (t1 – t2).

We express the mass of water m by the formula: m = ρ * V.

Qv = C * ρ * V * (t1 – t2).

Qw = 4200 J / kg * ºC * 1000 kg / m3 * 0.005 m3 * (20 ºC – 0 ºC) = 420,000 J.

Qw = 420,000 J <Ql = 510,000 J.

We see that the thermal energy that will be released when the water cools down is not enough to completely melt a piece of ice. The temperature in the bucket will be set to t2 = 0 ºC.

Answer: the temperature in the bucket will be set to t2 = 0 ºC.