00:01
To answer this question, the additional information that we need is the specific heat of aluminum and the specific heat of water.
00:13
So we need specific heats.
00:19
Then we'll make the assumption that all of the heat lost by the aluminum goes to the water.
00:32
I don't have access to your book, so i don't know what value it has for the specific heat of aluminum.
00:37
They seem to range from 0 .89 to 0 .91, a lot of them being right around 0 .9, so we'll assume that the specific heat of aluminum is 0 .902 joules per gram degrees celsius, while that of water is more commonly known to be 4 .184 joules per gram degree celsius.
01:07
Assuming the heat lost by the aluminum, so negative q of aluminum, equals the positive q of water as the water absorbs the heat that aluminum loses, then we can write a single equation where q of aluminum is equal to the specific heat of aluminum, .902 joules per gram degree celsius, multiplied by its mass, which we will be calculating, multiplied by its change in temperature, which will be its final temperature of 15, yeah, have thermal equilibrium temperatures 15 degrees celsius minus the initial temperature of 160 degrees celsius.
02:04
We set that equal to q for water which will be the specific heat of water 4 .184 joules per gram degrees celsius multiplied by the mass of the water which will be 80 grams multiplied by its change in temperature it goes to 15 degrees celsius from 0 degrees celsius but because this was ice and not water there's one other piece of information that we need to know and that's the enthalpy of fusion of water that's the the energy required to convert the solid ice into liquid water...