00:01
So in this example, we're told that we have 250 grams of coffee with the same heat capacity as water.
00:08
We're also told that our coffee is initially at 80 degrees celsius, and they're trying to add ice to the coffee in order to reduce the temperature to 60 degrees celsius.
00:17
So we can calculate our change in temperature of our coffee to be negative 20 kelvin.
00:23
And then we can calculate the change in temperature of our ice to be 60 degrees kelvin because they tell us that the original temperature, of our ice is going to be zero degrees celsius.
00:34
We also grabbed the heat effusion of water, which is 333 .7 joules per gram.
00:44
So in this question, they just want to know how much ice they have to add in order to reduce the heat of our coffee to 60 degrees celsius.
00:53
So we know that the transfer heat of our ice plus the transfer of heat of our coffee has to equal zero.
01:03
We're assuming that this is a closed system, we're not going to lose any heat or energy to our coffee cup or to our environment.
01:10
So the energy is conserved.
01:13
So we can also rewrite this.
01:15
It says the energy gained by our ice is going to be equal to the energy lost from our coffee...