HW 8.41

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Lei Wena Herme 1O
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Joined: Fri Jul 22, 2016 3:00 am

HW 8.41

Postby Lei Wena Herme 1O » Sun Jan 22, 2017 1:59 pm

A 50.0-g ice cube at 0.0C is added to a glass containing 400.0 g of water at 45.0C, What is the final temperature of the system? Assume that no heat is lost to the surroundings.

How would you go about this problem in setting up the equation(s) to find the final temperature?

Shruti Amin 1E
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Joined: Wed Sep 21, 2016 2:58 pm

Re: HW 8.41

Postby Shruti Amin 1E » Sun Jan 22, 2017 3:37 pm

You would have to do it in several steps.

Q(ice) = Q(ice1) + Q(ice2)
Q(ice) + Q(water) = 0

To find Q(ice1) : (50g/18.02g)*(6.01x10^3 J/mol)
Q(ice2): m*Cwater*T

Q(water) = m*Cwater*T

You'd plug in the values using the Appendixes and then you should get your answer.

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Re: HW 8.41

Postby Christina_F_3F » Sun Jan 22, 2017 5:27 pm

For this problem, I was curious why the grams of water used to calculate the heat of the water is 400g, rather than 450g? I understand that the reaction started with 400g of water, but by the time it is over, won't the 50g of ice have melted, resulting in 0g of ice and 450g of water in total?

Diego Zavala 2I
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Re: HW 8.41

Postby Diego Zavala 2I » Tue Jan 23, 2018 6:05 pm

The reason that you continue to calculate 400.0g of water and not 450.0g of water is that you are considering them two separate systems. You do this because the 400.0g of water have a higher temperature than the 50.0g of water at every time interval except when the two systems finally reach equilibrium.

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