00:02
This first part of the problem, the volume of the maximum volume of water in hot water heater at our home is v is equal to 25 liters.
00:30
Then its mass, in the second part of the problem, its mass will become volume into density.
00:46
As the volume is 25 liter and the density has been given as 1 .00 kilogram per liter.
00:58
So canceling this liter, the mass will come out to be 25 kilogram.
01:05
Initial temperature is 20 degrees celsius or 20 plus 273 kelvin.
01:15
And final temperature up to which the temperature of water has to be raised is 45 degrees celsius or this is 45 plus 273 kelvin so the change of temperature will be t2 minus t1 so if we subtract these two temperatures 273 will be subtracted will be cancelled out so the remaining temperature difference will be 25 kelvin.
01:49
Then the specific heat capacity of water here is given as 4 ,190 juice per kilogram per kelvin.
02:05
So using the expression for heat absorbed by the water, which is given as delta q is equal to m into c into delta t.
02:15
This heat absorbed here by the water will be given by mass which is 25.
02:20
Kilogram, specificate capacity 4190, joules per kilogram per kelvin, multiplied by change of temperature, which is 25 kelvin.
02:36
So canceling the similar units, kelvin with kelvin, kg with kg.
02:42
Here, this delta q comes out to be 2 .62 into 10 dash to the power 6 jou...