0*334 = 7014JEnergy removed colling ice = 21.0*2.2 * (3-0) = 138.6J (MUST be a positive quantity)Total = 7416JThis doesnt match any of your values either! I think the answers are all wrong!Edit: The specific heat capacity of ice is nearer to 2.......
I would do it differently. Here's my working using only positive quantities:
Energy removed cooling water = 21.0*4.18*(3-0) =263.34J (MUST be a positive quantity)
Energy removed freezing water =21.0*334 = 7014J
Energy removed colling ice = 21.0*2.2 * (3-0) = 138.6J (MUST be a positive quantity)
Total = 7416J
This doesn't match any of your values either! I think the answers are all wrong!
Edit: The specific heat capacity of ice is nearer to 2.02 than 2.2. Check the values you were given.
1) You are right, in that the question does ask for the heat capacity of the sample. Seems to me that the teacher ( or question writer ) is being a bit loose about their use of terminology.
But what is important is that you have read the question accurately and have shown that you understand what is being asked about. In the long run it doesn't actually matter what the teacher thinks ... you have it right.... move on.
2) You have this all right until you get to the maths bit.
The sample is cooling down... it is losing heat at each of the three steps you identify. So the total loss has to be just the three bits added together. You seem to have decided that some of the losses are negative , meaning gains in heat !!
263 + 7014 + 139 = 7416 J
quite close to (c) though how they get that number for (c) I don't know as the figures definitely produce 7400 J as correct. I wonder if there is some more teacher slippage here?
