8.41 6th edition

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Madeline Ho 1C
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Joined: Mon Apr 23, 2018 3:00 am

8.41 6th edition

Postby Madeline Ho 1C » Mon Feb 04, 2019 12:45 am

A 50.0-g ice cube at 0.0 C is added to a glass containing 400.0 g of water at 45.0 C. What is the final temperature of the system (see Tables 8.2 and 8.3)? Assume that no heat is lost to the surroundings.
For this question, I found the deltaH for the phase change from ice to water, the heat absorbed by the melted ice, and the heat released by water. Then I set up my equation
deltaH + heat absorbed by melted ice = -heat released by water. I got 62°C but the answer in the textbook is 31°C. Can someone help me with this? Thanks.

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Re: 8.41 6th edition

Postby Tam To 1B » Mon Feb 04, 2019 10:26 am

In this problem, you add the enthalpy of fusion, heat absorbed by ice, and heat released by water.
In order to find the enthalpy of fusion, you convert 50.0 g H20(s) to mol and multiply it by the enthalpy of fusion. This would be 2.775 mol H20 * 6.01E3 J/mol (convert kJ in textbook to J).
The heat absorbed by ice would be 50.0 g * 4.184 J/gC * (T-0 C) since it's added to the water.
The heat released by water would be 400 g * 4.184 J/gC * (T-45 C).

Add all these values and solve for T to get 31 C.

Laura Gong 3H
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Joined: Fri Sep 28, 2018 12:26 am
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Re: 8.41 6th edition

Postby Laura Gong 3H » Mon Feb 04, 2019 2:18 pm

I used a similar method to answer this question.
Using the First Law of Thermodynamics, I know that q(ice)=-q(water) where q(ice) is equal to the enthalpy of the phase change of ice (q=n*deltaH(fus)) and the enthalpy of the unknown temperature change (q=50.0*4.184*(Tf-0)). q(water) is equal to the enthalpy of the unknown temperature change (q=400.0*4.184*(Tf-45)). Once I set up that full equation, I solved for Tf.

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