# It was decided to throw a brass ball with a temperature of -50 degrees C into water at a temperature of 25

It was decided to throw a brass ball with a temperature of -50 degrees C into water at a temperature of 25 degrees C so that the water would completely freeze. What is the minimum mass of a ball if the mass of water is 200g? Is this experiment possible in practice?

To find the minimum mass of the taken brass ball, we use the equality: Qsh = Qw; Sl * msh * (0 – tsh) = (Sv * (tv – 0) + λw) * mw, whence we express: mw = (Sv * tv + λw) * mw / (Sl * (-tw)).

Const: St. – beats. heat capacity of water (Sv = 4.18 * 10 ^ 3 J / (kg * K)); λw – beats heat of crystallization of water (λw = 332.4 * 10 ^ 3 J / kg); Sl – beats. heat capacity of brass (Cl = 920 J / (kg * K)).

Data: tv – initial water temperature (tv = 25 ºС); mw is the mass of water (mw = 200 g, in the SI system mw = 0.2 kg); tsh – initial temperature of the brass ball (tsh = -50 ºС).

Let’s perform the calculation: mw = (Sv * tv + λw) * mw / (Sl * (-tw)) = (4.18 * 10 ^ 3 * 25 + 332.4 * 10 ^ 3) * 0.2 / (920 * (-50)) = 1.9 kg.

Experiment Opportunity:

– the diameter of the brass ball: d = (6V / Π) ^ 1/3 = (6 * mw / (ρl * Π)) ^ 1/3 = (6 * 1.9 / (8500 * 3.14)) ^ 1 / 3 = 0.0753 m = 7.53 cm.

– the height of the liquid column: h = V / S = 4mw / (ρw * Π * d ^ 2) = 4 * 0.2 / (1000 * 3.14 * (0.0753) ^ 2) = 0.0449 m = 4.49 cm.

Water will not completely cover the brass ball, therefore, heat exchange of the ball with the environment will occur, and the experiment will be possible only in an isolated system.

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