00:01
So we want to calculate the minimum amount of electrical energy required by the freezer in order to freeze 1 .2 kilograms of water initially at 20 degrees celsius, where the freezer's temperature is minus 20 degrees celsius.
00:20
And so the first thing we need to calculate is how much heat flow out of the water does it take to bring 20 degrees celsius of water to minus 20 degrees celsius.
00:33
And so we can go ahead and calculate this right away as q is equal to minus the mass of the water we have, multiplied by the specific heat of the water, multiplied by the change in temperature of the water.
00:48
And note that i am putting it in absolute value because i am indicating the direction of heat flow by hand with the minus sign up front, minus the amount of energy it takes to change the phase of water from a liquid to a solid, and then minus the amount of heat it takes to decrease the temperature of the ice from 0 degrees celsius to minus 20 degrees celsius.
01:20
And note that we need this separate term because the heat capacity, or the specific heat, i should say, of ice, is different from that of water.
01:32
And so be careful of that.
01:34
And so this expression for the amount of heat to bring 20 degrees of celsius water to minus 20 degrees of ice can be factored slightly where we can bring the negative sign in the m out front and we have change.
01:49
The specific heat of water multiplied by the change of temperature of the water minus the latent heat of fusion minus c -5 i multiplied by delta t i and so now we can plug in all of our values and we have negative 1 .2 kilograms multiplied by the specific heat of water in kilojoules per kilogram per kelvin multiplied by 20 kelvin the change in temperature of the water plus the latent heat of fusion of 333 .7 kilojoules per kilogram and now i'm just going to move down a bit plus 2 .1 kilojoules per kilogram per kelvin the specific heat of ice multiplied by the change of temperature of the ice and calculating this we obtain negative 551 .304 kilojoules...