Solid Copper Bowl
Solid Copper Bowl
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University Physics - The Absorption of Heat by Solids and Liquids?
A 1.50 g copper bowl contains 220 g of water, both at 20 degrees celsius. A very hot 300g copper cylinder is dropped into the water causing the water to boil, with 5 g being converted to steam. The final temperature of the system is 100 degrees celsius. Neglect energy transfers with the environment.
What is the original temperature of the cylinder?
I know to find this, I need to find the total heat transferred to the water (Qw) by summing specific heat and latent heat, also the heat transfered to the bowl of copper (Qc). And in order to find the original temperature of the cylinder I would use
Qw + Qc = m (specific heat) (Ti - Tf)
what i dont understand is why take the difference of the initial temperature and final temperature. when change in temperature is usually (Tf - Ti) the difference between the final temperature and the initial temperature.
why is this so?
mc(Tf - Ti) + Qw + Qc = 0
Qw+Qc = -mc(Tf-Ti)
Qw+Qc = mc(Ti-Tf)
You skipped two steps.
The cylinder loses heat, whereas water and bowl gain heat. So the amount of heat gained by water and bowl equals to the amount of heat the cylinder loses. It makes perfect sense.