## anonymous one year ago 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?

1. UnkleRhaukus

2. anonymous

160g

3. UnkleRhaukus

158.2 g

4. UnkleRhaukus

We can remember that density $$\rho = m/V$$ Solve this equation for the mass $$m$$, then plug in the values for $$\varrho$$, and $$V$$

5. anonymous

So 160 * 1.00?

6. UnkleRhaukus

yes

7. anonymous

so 160 g

8. UnkleRhaukus

yes 158.2 g

9. UnkleRhaukus

NB ( 1 cm^3 = 1 ml )

10. anonymous

what does the NB stand for?

11. UnkleRhaukus

note well: $$1\,[\text{cm}^3] = 1\,[\text{mL}]$$)

12. UnkleRhaukus

13. anonymous

Oh. Thanks.

14. UnkleRhaukus

OK, so you have (a), what do you think about (b)?

15. anonymous

would temperatures need to be added?

16. UnkleRhaukus

is temperature an extensive quantity, or intensive ?

17. anonymous

intensive

18. UnkleRhaukus

right, so the temperature should become the average

19. UnkleRhaukus

(not the sum)

20. UnkleRhaukus

what is the average of 17°C and 17°C ?

21. anonymous

17

22. UnkleRhaukus

* is suppose it really should be the weighted average, but yes the temperatures are the same so the temperature wont change

23. UnkleRhaukus

makes sense?

24. anonymous

not exactly

25. UnkleRhaukus

which bits are unclear?

26. anonymous

how I am suppose to use the average to figure out the final temperature

27. UnkleRhaukus

i don't think you are expected calculate the temperature, the temperatures are the same

28. anonymous

so it would be 17 as the final?

29. 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°

30. anonymous

Alright. That makes sense now.

31. anonymous

Thanks.

32. UnkleRhaukus

If they were different temperatures, then the final temperature would be some in-between

33. anonymous

Gotcha. It makes sense now.

34. UnkleRhaukus

so what do you got for (c) ?

35. anonymous

You would multiple the final mass by the density to get volume.

36. anonymous

which gives 160

37. anonymous

So 160/160 = 1.00g/mL

38. 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

39. 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.

40. UnkleRhaukus

you don't have you multiply or divide to get (c),

41. anonymous

why would we need to know the temperature for the final density?

42. UnkleRhaukus

because the density of water is temperature dependent

43. anonymous

Okay. Because of the intensive and extensive

44. anonymous

So you only have to look at it? Not do any math?

45. 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]

46. anonymous

what makes you realize that?

47. UnkleRhaukus

it's given in the question

48. UnkleRhaukus

we found the final temperature in (b), and the question tell us the density of water at this temperature

49. anonymous

Yeah.

50. 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

51. 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.

52. anonymous

because intensive is independent?

53. UnkleRhaukus

Yes, intensive properties are independent of the amount we have

54. anonymous

That makes sense now.

55. anonymous

Thank you again for the help.