## Practice Midterm Q3A Winter 2013

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

William_Lee_Dis_3H
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### Practice Midterm Q3A Winter 2013

When calculating the work done by the steam, which mass of the water should we use from part A? The mass of the water was rounded to 77.0 grams but then in the solutions they used the non-rounded value to calculate the work done. This results in answers differing by about 5 Joules.

Chem_Mod
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### Re: Practice Midterm Q3A Winter 2013

The answer does differ by about 5 Joules, but w is -13 kJ. 5 J is a very small portion of it thus could be ignored. In your calculations, you should use more accurate data and take sig figs at the last step in order to decrease error.

Coco Hailey 2E
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### Re: Practice Midterm Q3A Winter 2013

The answers explain that q_katana = (1.45kg)(473 J/kg*K)(289˙C). Why isn't 289˙C converted to Kelvin?

Wendy_Sandoval1k
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### Re: Practice Midterm Q3A Winter 2013

The answer is actually explained as q_katana= (1.45kg)(473 J/kg*K)(25.0˙C- 314.0˙C) and they didn't convert into Kelvin because they are looking for the difference and it would be the same answer if it is in celsius or kelvin, but I feel like it should be converted just to be able to cancel out with the kelvin in (473 J/kg*K).

Carolyn Huh 1K
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### Re: Practice Midterm Q3A Winter 2013

The answer should be rounded at the last step according to sig figs. Otherwise, it is very likely to get a different answer.

Simone Seliger 1C
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### Re: Practice Midterm Q3A Winter 2013

Wendy_Sandoval1k wrote:The answer is actually explained as q_katana= (1.45kg)(473 J/kg*K)(25.0˙C- 314.0˙C) and they didn't convert into Kelvin because they are looking for the difference and it would be the same answer if it is in celsius or kelvin, but I feel like it should be converted just to be able to cancel out with the kelvin in (473 J/kg*K).

I have the same concern. Would we be marked down for correctly converting it and getting a different value because our temp. would be in K?