A student pours 46.8g of water at 17oC into a beaker containing 111.4g or water at 17oC. The density of water at 17oC is 1.00 g/mL. (a) What is the final mass? (b) What is the final temperature? (c) What is the final density?

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A student pours 46.8g of water at 17oC into a beaker containing 111.4g or water at 17oC. The density of water at 17oC is 1.00 g/mL. (a) What is the final mass? (b) What is the final temperature? (c) What is the final density?

Mathematics
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Add the volumes
160g
158.2 g

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Other answers:

We can remember that density \(\rho = m/V\) Solve this equation for the mass \(m\), then plug in the values for \(\varrho\), and \(V\)
So 160 * 1.00?
yes
so 160 g
yes 158.2 g
NB ( 1 cm^3 = 1 ml )
what does the NB stand for?
note well: \( 1\,[\text{cm}^3] = 1\,[\text{mL}]\))
Oh. Thanks.
OK, so you have (a), what do you think about (b)?
would temperatures need to be added?
is temperature an extensive quantity, or intensive ?
intensive
right, so the temperature should become the average
(not the sum)
what is the average of 17°C and 17°C ?
17
* is suppose it really should be the weighted average, but yes the temperatures are the same so the temperature wont change
makes sense?
not exactly
which bits are unclear?
how I am suppose to use the average to figure out the final temperature
i don't think you are expected calculate the temperature, the temperatures are the same
so it would be 17 as the final?
yeah, if you mix so water that is 17°C, with some more water that is 17°C, the temperature of the mixture will be 17°
Alright. That makes sense now.
Thanks.
If they were different temperatures, then the final temperature would be some in-between
Gotcha. It makes sense now.
so what do you got for (c) ?
You would multiple the final mass by the density to get volume.
which gives 160
So 160/160 = 1.00g/mL
(c) is asking for the final density, we know that density (like temperature, but unlike mass) is an intensive quality, i.e. density is a property of the material, rather than the object
the question states The density of water at 17°C is 1.00 [g/mL]. We found the final temperature of the mixture to be 17°C.
you don't have you multiply or divide to get (c),
why would we need to know the temperature for the final density?
because the density of water is temperature dependent
Okay. Because of the intensive and extensive
So you only have to look at it? Not do any math?
right , there is no math for (c), it is just: notice that our final temperature 17°C, is one at which we know the density to be 1 [g/mL]
what makes you realize that?
it's given in the question
we found the final temperature in (b), and the question tell us the density of water at this temperature
Yeah.
if the temperatures were different, and we calculated a weighted average temperature, we might use a table to look up the density of water at the temperature
We are combining objects with the same intensive properties, whereas, the extensive property (the mass) is the sum the intensive properties (temperature and density) of the mixture will be equal.
because intensive is independent?
Yes, intensive properties are independent of the amount we have
That makes sense now.
Thank you again for the help.

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