Hot metal added to Water, Heat Capacity Problem

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SUMMARY

The discussion focuses on calculating the final temperature of a copper-water mixture, where a 100.0 g copper sample at 100°C is mixed with 50.0 g of water at 26.5°C. The specific heat of copper is established as 0.385 J/g K. The heat transfer equation, q = q(copper) + q(water), is utilized, emphasizing that the heat lost by copper equals the heat gained by water. The final temperature can be determined by setting the heat loss of copper equal to the heat gain of water, leading to a solvable equation for Tfinal.

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Homework Statement



A 100.0 g copper sample (specific heat 0.385 J/g K) at 100 C is added to 50.0 g of water at 26.5 C. What is the final temperature of the copper-water mixture?

Homework Equations



q = q(copper) + q(water)

q = (heat capacity)(change in Temp.)

q = (specific heat)(mass)(change in Temp.)


The Attempt at a Solution



So far i have found the heat capacity of copper to be 0.385 J/K

I know the metal loss of heat must equal the waters gain in heat but i do not know how to calculate the gain/loss.
 
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It seems you have the idea. To find the final temperature, try setting -qcopper = (specific heat)(mass)(change in Temp.) equal to qwater = (specific heat)(mass)(change in Temp.)

Another thing to keep in mind that should help you solve for final temperature, is that the at the end, the temperature of the water and the copper are going to be equal. So, if you think about what (change in Temp) really is, you should find yourself with an equation and one unknown, Tfinal, and from there you should be able to solve for temperature at the end.
 

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