# Homework Help: Specific heat capacity problem

1. Feb 12, 2012

### lionely

A bath contains 100kg of water at 60C. Hot and cold taps are then turned on to deliver 20kg per minute each at temperatures of 70C and 10C respectively. How long will it be before the temperature in the bath has dropped to 45C? Assume complete mixing of the water and ignore heat losses.

I've never done a specific heat capacity question that involves time so I'm not sure on how to even start.

2. Feb 12, 2012

### tiny-tim

hi lionely!

general strategy …

start by giving everything a letter, and then translate the question into equations​

in this case, call the amount of added hot water "h", the amount of added cold water "c", the time "t", and then write out how h and c depend on t

finally apply newton's law of cooling …

what do you get?

3. Feb 12, 2012

### lionely

I don't know what Newton's law of cooling is I'm only in the 11th grade :(

But umm is it

h + c/t?

4. Feb 12, 2012

### tiny-tim

??

what rule have you been taught for finding the final temperature of a mixture?

5. Feb 12, 2012

### lionely

Ummm the basic

heat given out = heat received

mass x specific heat capacity x temp change = mass specific heat capacity x temp change.

6. Feb 12, 2012

### tiny-tim

hmm … i don't like that

i] it skates over the fact that one of the changes is negative

ii] it doesn't help if (as here) there's three masses​

better would be …

∑ (mass x specific heat capacity x temp change) = 0

try that!​

7. Feb 12, 2012

### lionely

what is the temp change add all the temps and average and just ignore the time ?

8. Feb 12, 2012

### tiny-tim

yes, ignore the time

(though you will need the time to calculate how much water is being added )

9. Feb 12, 2012

### lionely

so for the temp change (60 + 70 + 10/3)-45?

10. Feb 12, 2012

### tiny-tim

i've no idea what you're doing

11. Feb 12, 2012

### lionely

∑ (mass x specific heat capacity x temp change) = 0 I was trying to follow what you said.

Is the total mass like the h + c + the 100kg water?

12. Feb 12, 2012

### tiny-tim

that's later

first

13. Feb 12, 2012

### lionely

cold = c
time = t

h = 20kg/60 seconds? Every minute another 20kg is added same for the cold.

That's how h and c depend on t.

14. Feb 12, 2012

### technician

The original 100kg of water at 60C has to be cooled to 45C.... Can you calculate how much energy must be removed from this 100kg of water?
This heat energy must go to warm up the mixture of (20kg at 70C + 20kg at 10C) arriving per second. This mixture ends up at 45C.
Does this help you to see what you need to sort out......ifr you had 40kg arriving per second ,what temp would it be to have the same effect as the 2 lots of 20kg?

15. Feb 12, 2012

### tiny-tim

not quite

you need h on the left and a function of t on the right

(ie it needs to have t in it !)

(same for c, of course)

16. Feb 12, 2012

### lionely

I'm sorry but I'm so confused is the energy needed to be removed for the 60 C water to reach 45 C = 100kg x 4200 x (60c - 45c) = 6,300,000 J?

17. Feb 12, 2012

### technician

Yes !!!and that energy goes to warm up the mixture to 45C. Can you see what the temp of the mixture is as it enters the bath?

18. Feb 12, 2012

### lionely

is it 20kg of water at 70c + 20kg of water at 10c /2 ? so 40c?

19. Feb 12, 2012

### technician

YES so 40kg water at 40C per MINUTE needs to be warmed to 45C.... how much energy is needed to do this.... how many minutes?

20. Feb 12, 2012

### lionely

Omg thank you it's 6,300,000/840,000 = 7.5 minutes I don't know how I couldn't see that before. So stupid...

21. Feb 12, 2012

### technician

that is what I got 7.5 minutes

22. Feb 12, 2012

### lionely

Yeah that's what it says in the back of my book.

23. Feb 12, 2012

### technician

well done..it is not Newton's law of cooling !! it is straightforward heat lost = heat gained
Newton's law of cooling is something completely different and you have probably not met it yet

Last edited: Feb 12, 2012