anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
Add the volumes
anonymous
  • anonymous
160g
UnkleRhaukus
  • UnkleRhaukus
158.2 g

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UnkleRhaukus
  • UnkleRhaukus
We can remember that density \(\rho = m/V\) Solve this equation for the mass \(m\), then plug in the values for \(\varrho\), and \(V\)
anonymous
  • anonymous
So 160 * 1.00?
UnkleRhaukus
  • UnkleRhaukus
yes
anonymous
  • anonymous
so 160 g
UnkleRhaukus
  • UnkleRhaukus
yes 158.2 g
UnkleRhaukus
  • UnkleRhaukus
NB ( 1 cm^3 = 1 ml )
anonymous
  • anonymous
what does the NB stand for?
UnkleRhaukus
  • UnkleRhaukus
note well: \( 1\,[\text{cm}^3] = 1\,[\text{mL}]\))
UnkleRhaukus
  • UnkleRhaukus
anonymous
  • anonymous
Oh. Thanks.
UnkleRhaukus
  • UnkleRhaukus
OK, so you have (a), what do you think about (b)?
anonymous
  • anonymous
would temperatures need to be added?
UnkleRhaukus
  • UnkleRhaukus
is temperature an extensive quantity, or intensive ?
anonymous
  • anonymous
intensive
UnkleRhaukus
  • UnkleRhaukus
right, so the temperature should become the average
UnkleRhaukus
  • UnkleRhaukus
(not the sum)
UnkleRhaukus
  • UnkleRhaukus
what is the average of 17°C and 17°C ?
anonymous
  • anonymous
17
UnkleRhaukus
  • UnkleRhaukus
* is suppose it really should be the weighted average, but yes the temperatures are the same so the temperature wont change
UnkleRhaukus
  • UnkleRhaukus
makes sense?
anonymous
  • anonymous
not exactly
UnkleRhaukus
  • UnkleRhaukus
which bits are unclear?
anonymous
  • anonymous
how I am suppose to use the average to figure out the final temperature
UnkleRhaukus
  • UnkleRhaukus
i don't think you are expected calculate the temperature, the temperatures are the same
anonymous
  • anonymous
so it would be 17 as the final?
UnkleRhaukus
  • UnkleRhaukus
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°
anonymous
  • anonymous
Alright. That makes sense now.
anonymous
  • anonymous
Thanks.
UnkleRhaukus
  • UnkleRhaukus
If they were different temperatures, then the final temperature would be some in-between
anonymous
  • anonymous
Gotcha. It makes sense now.
UnkleRhaukus
  • UnkleRhaukus
so what do you got for (c) ?
anonymous
  • anonymous
You would multiple the final mass by the density to get volume.
anonymous
  • anonymous
which gives 160
anonymous
  • anonymous
So 160/160 = 1.00g/mL
UnkleRhaukus
  • UnkleRhaukus
(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
UnkleRhaukus
  • UnkleRhaukus
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.
UnkleRhaukus
  • UnkleRhaukus
you don't have you multiply or divide to get (c),
anonymous
  • anonymous
why would we need to know the temperature for the final density?
UnkleRhaukus
  • UnkleRhaukus
because the density of water is temperature dependent
anonymous
  • anonymous
Okay. Because of the intensive and extensive
anonymous
  • anonymous
So you only have to look at it? Not do any math?
UnkleRhaukus
  • UnkleRhaukus
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]
anonymous
  • anonymous
what makes you realize that?
UnkleRhaukus
  • UnkleRhaukus
it's given in the question
UnkleRhaukus
  • UnkleRhaukus
we found the final temperature in (b), and the question tell us the density of water at this temperature
anonymous
  • anonymous
Yeah.
UnkleRhaukus
  • UnkleRhaukus
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
UnkleRhaukus
  • UnkleRhaukus
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.
anonymous
  • anonymous
because intensive is independent?
UnkleRhaukus
  • UnkleRhaukus
Yes, intensive properties are independent of the amount we have
anonymous
  • anonymous
That makes sense now.
anonymous
  • anonymous
Thank you again for the help.

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