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### HW 8.25

Posted: **Sat Jan 09, 2016 2:44 pm**

by **504594108**

I do not understand how the Solutions Manuel solves this problem because it's asking for the internal energy but the Solutions Manuel uses heat capacity and heat formulas. Can someone please explain how it was solved?

### Re: HW 8.25

Posted: **Sat Jan 09, 2016 7:38 pm**

by **Michelle Pham 1G**

Heat capacity is linked to internal energy because heat capacity is equal to heat supplied/temperature change, which in equation form is C= q/delta T

Change in internal energy (delta U) is equal to q (heat) + w (work)

Because the problem specifies that it is a constant-volume calorimeter, w= 0 in this case. That means that the change in internal energy (delta U) is equal to q (heat).

To solve this problem, you must first find the 1. heat capacity of the calorimeter (Ccal), and then use that information to find 2. the energy change (q). This will then give you change in internal energy.

### Re: HW 8.25

Posted: **Mon Jan 18, 2016 4:41 pm**

by **Ajith Raja 2L**

On the solutions manual, the change in temperature that is used is given in Kelvin. For example, they use 7.32K. However, in the problem, the temperature is given in Celsius. Don't we have to convert Celsius to Kelvin before we continue with the problem? If so, why is it not done in the solutions manual? Is this a typo?...

### Re: HW 8.25

Posted: **Tue Jan 19, 2016 9:21 pm**

by **Michelle Pham 1G**

It actually does not matter whether Celsius or K is used because a change of 1 degree Celsius is the same as a change of 1 K. This is because K= C + 273. The numbers differ by 273, but if there is any change in C, the same change will occur in K.

### Re: HW 8.25

Posted: **Sun Jan 24, 2016 9:09 pm**

by **Aishwarya Raich**

Why is this equation not factoring in mass? The solution manual uses the equation C= q/ delta T, however the equation that we are provided with is originally q=nc(delta T). Aren't the number of moles an important factor to consider in these types of problems?