HW Problem 4A.9

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Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

HW Problem 4A.9

Postby Gisela F Ramirez 2H » Wed Jan 30, 2019 9:54 pm

A piece of copper of mass 20.0 g at 100.0 degrees Celsius is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 degrees Celsius. Calculate the final temperature of water. Assume that no energy is lost to surroundings.

How do I go about solving this problem?

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

Re: HW Problem 4A.9

Postby Vincent Li 4L » Wed Jan 30, 2019 9:59 pm

For this problem, think about where the heat is going. Because there is a temperature difference, energy will be transferred from the copper and into the water. Since heat released = - heat absorbed, we can use q = mC for both substances, and we solve for Tfinal since we should know all other values. Hope this helps.

Tuong-Minh Tran 1C
Posts: 30
Joined: Fri Sep 28, 2018 12:19 am

Re: HW Problem 4A.9

Postby Tuong-Minh Tran 1C » Wed Jan 30, 2019 10:03 pm

The most important concept to take into account here is that the energy it takes to heat up the water to its final temperature is equal to the amount of heat energy the copper loses. It can be portrayed as this:

ΔH(copper)=-ΔH(water)

After that, we refer to the general equation ΔH=m*C(specific)*ΔT and solve for the final temperature of water:

(20.0)(0.38)(T(final)-100.0)=-(50.7)(4.18)(T(f)-22.0)

Hope that helps!

Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: HW Problem 4A.9

Postby Gisela F Ramirez 2H » Wed Jan 30, 2019 10:51 pm

Thank you so much!


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