How many degrees will the boiling water cool (at 100 ° C), which completely fills a drinking tank with a capacity of 25 liters

How many degrees will the boiling water cool (at 100 ° C), which completely fills a drinking tank with a capacity of 25 liters, if this boiling water gives up to the surrounding space an amount of heat equal to 1425 kJ? The answer should be 13.6 ° C.

Given:

t1 = 100 ° Celsius – the initial temperature of boiling water;

V = 25 liters = 0.025 m3 (cubic meters) – the volume of the drinking tank;

Q = 1425 kJ = 1425000 Joules – the amount of energy given off by boiling water into space;

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

ro = 1000 kg / m3 (kilogram per cubic meter) – water density.

It is required to determine dt (degree Celsius) – by how many degrees the temperature of the boiling water will drop.

Let’s find the mass of boiling water:

m = ro * V = 1000 * 0.025 = 25 kilograms.

Then the temperature of the boiling water will drop by:

dt = Q / (c * m) = 1425000 / (4200 * 25) = 1425000/105000 = 1425/105 = 13.6 ° Celsius (the result has been rounded to one decimal place).

Answer: the temperature of the boiling water will drop by 13.6 ° Celsius.