A piece of iron weighing 2 kg with t500 degrees was thrown into a tank containing water with a mass

A piece of iron weighing 2 kg with t500 degrees was thrown into a tank containing water with a mass of 10 kg at t 20 degrees. Moreover, a certain amount of water turned into steam. Final t in the tank = 24 degrees. Determine the mass of the evaporated water.

Given:

m1 = 10 kilograms is the mass of water in the tank;

t1 = 20 ° Celsius – initial water temperature;

c1 = 4200 J / (kg * C) – specific heat of water;

q = 2260000 Joule / kilogram – specific heat of vaporization of water;

m2 = 2 kilograms – the mass of a piece of iron thrown into water;

t2 = 500 ° Celsius – temperature of a piece of iron;

c2 = 444 J / (kg * C) – specific heat of iron;

t = 24 ° Celsius – steady-state temperature in the tank.

It is required to determine dm (kilogram) – the mass of the evaporated water.

Let’s find the amount of heat that was released as a result of cooling a piece of iron:

Q = c2 * m2 * (t2 – t) = 444 * 2 * (500 – 24) = 888 * 476 = 422688 Joules.

This heat was spent on heating and evaporating the water. Let the amount of evaporated water be dm. Then:

Q = c1 * (m1 – dm) * (t – t1) + c1 * dm * (100 – t1) + q * dm, where 100 is the boiling point of water;

Substitute the known numerical values:

422688 = 4200 * (10 – dm) * (24 – 20) + 4200 * dm * (100 – 20) + 2260000 * dm;

422688 = 16800 * (10 – dm) + 336000 * dm + 2260000 * dm;

422688 = 168000 – 16800 * dm + 336000 * dm + 2260000 * dm;

422688 – 168000 = 2579200 * dm;

254688 = 2579200 * dm;

dm = 254688/2579200 = 0.1 kilograms (the result has been rounded to the nearest one hundredth).

Answer: the mass of the evaporated water is 0.1 kilograms.