00:01
So this question, we have a 150 gram sample of a metal at 75 degrees celsius, is added to 150 grams of water at 15 degrees celsius.
00:11
The temperature of the water rises to 18 .3.
00:15
Calculate the specific heat capacity in the metal, assuming that all the heat loss by the metal is gained by the water.
00:22
So this one is a little tricky.
00:23
It's still a calerometry problem, so we're still going to set heat gained equals heat loss.
00:29
But you have to understand that the temperature of the water, if it ends at 18 .3, that also means that the metal is ending at 18 .3.
00:39
So that means that we actually can do this in two separate steps.
00:46
So we're first going to calculate the heat gained by the water.
01:09
So we've got q equals mc delta t.
01:14
So we're going to figure out what is q for the water.
01:19
So the water, we had 150 grams of water.
01:24
The specific key is 4 .18 joules per gram degree celsius...