## 8.41 6th edition

$\Delta U=q+w$

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Joined: Mon Apr 23, 2018 3:00 am

### 8.41 6th edition

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

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

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.