An aluminum body with a mass of 0.48 kg and a temperature of 240 C was lowered into water

An aluminum body with a mass of 0.48 kg and a temperature of 240 C was lowered into water weighing 3.68 kg at a temperature of 22 C. Find the steady-state temperature 0.

Given:

m1 = 0.48 kilograms – the mass of the aluminum body, which was lowered into the water;

t1 = 240 ° degrees Celsius – the temperature of the aluminum body;

m2 = 3.68 kilograms – the mass of water;

t2 = 22 ° Celsius – water temperature;

c1 = 900 J / (kg * C) – specific heat capacity of aluminum;

c2 = 4200 J / (kg * C) – specific heat capacity of water.

It is required to determine t (degree Celsius) – the steady-state temperature.

According to the law of thermodynamics, the amount of heat that an aluminum body will give up during cooling will be equal to the amount of heat received by water for heating, that is:

Q1 = Q2;

c1 * m1 * dt1 = c2 * m2 * dt2;

c1 * m1 * (t1 – t) = c2 * m2 * (t – t2);

c1 * m1 * t1 – c1 * m2 * t = c2 * m2 * t – c2 * m2 * t2;

c1 * m1 * t1 + c2 * m2 * t2 = c2 * m2 * t + c1 * m1 * t;

c1 * m1 * t1 + c2 * m2 * t2 = t * (c2 * m2 + c1 * m1);

t = (c1 * m1 * t1 + c2 * m2 * t2) / (c2 * m2 + c1 * m1) =

= (900 * 0.48 * 240 + 4200 * 3.68 * 22) / (4200 * 3.68 + 900 * 0.48) =

= (103680 + 340032) / (15456 + 432) = 443712/15888 = 27.9 ° Celsius (the result has been rounded to one decimal place).

Answer: The steady-state temperature will be 27.9 degrees Celsius.