21.2 Câ
There are a few assumptions to note before solving this problem. First assumption is that the aluminum cup is at the same temperature as the water at the start. The second assumption is that the from the copper block is transferred slowly enough that no water is lost due to boiling or evaporation, or that such loss is insignificant.
Since the temperatures are in a range where there will be no phase changes, we can make the problem easier by arbitrary considering our starting temperature to be 16.0 âC for everything. So we'll consider the starting energy for the aluminum and water to be 0. So let's calculate the amount of excess energy the copper block has over the base temperature of 16.
235 - 16 = 216
216 * 390 J/kgâ‹…Câ * 0.225 kg = 19217.25 J
We now have 19217.25 Joules of energy with which to raise the temperature of 225 grams of copper, 155 grams of aluminum, and 835 grams of water. Let's see how much energy is needed to raise the temperature of all 3 items by 1 degree.
0.225 Kg * 390 J/kgâ‹…Câ + 0.155 kg * 900 J/kgâ‹…Câ + 835 * 4186 J/kgâ‹…Câ
= 87.75 J/Câ + 139.5 J/Câ + 3495.31 J/Câ
= 3722.56 J/Câ
Now we have a simple matter of multiplication to get the temperature rise from our starting point.
19217.25 J / 3722.56 J/Câ = 5.162374817 Câ
So the temperature rises by 5.162374817 Câ from our starting point. Just add that to the starting point:
16.0 Câ + 5.162374817 Câ = 21.16237482 Câ
Finally, round to 3 significant figures, giving 21.2 Câ
