00:01
In this problem, a number of aluminium balls are to be quenched in a water bath at a specified rate.
00:07
We need to calculate the rate at which the heat needs to be removed from the water in order to keep its temperature constant.
00:13
We assume that the thermal properties of the balls are constant.
00:16
The balls are at a uniform temperature before and after quenching, and the changes in kinetic and potential energy are negligible.
00:23
Now the properties of the aluminium balls are at the average temperature which is the temperature at the middle of the initial and final temperatures which is 370 kelvin.
00:37
We have the density of the aluminium balls to be 2 ,700 kg per cubic metre and the specific heat c to be 0 .97 kilojoules per kg degrees celsius.
00:50
Now we'll take a single ball as the system.
00:56
So we can write the energy balance equation for this closed system as e in minus e out is equal to the change in the energy of the system, delta e.
01:09
Now we only have energy leaving the system in the form of heat as the balls lose heat to the water.
01:20
And this is equal to a change in the internal energy.
01:23
Of the ball delta u.
01:27
So we can write this as m c t1 minus or t2 minus t1 and this is because we can write delta u as m into u2 minus u1 the mass specific change in internal energies and we can write that as mc t2 minus t1.
01:58
So if we multiply both sides by minus one, we get that the heat out of the ball and into the water q out is equal to mct1 minus t2.
02:14
So we multiply both sides by minus 1.
02:18
Now in order to be able to use the energy balance equation, we first need to find the mass of the element in ball...