Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

You add an unknown volume of milk of $5.2 ^\circ C$ to a cup of coffee ($40 mL$ of water, temperature: $80.3 ^\circ C$). After a while of stirring the temperature reaches $73.2 ^\circ C$. The properties of water are the same as those of coffeewater. Neglect the warmth capacity of the cup:

  • Calculate the warmth that has left the coffee.

  • Calculate the added milk in mL.

$$Q = cm\Delta T$$

$$ c = 4.18\times10^3$$

$$m = 0.04 $$

(I believe so, 1L of water is 1kg, so 40ml=0,04 kg)

$$\Delta T = -7.1$$

$$Q=0.04\times4.18\times10^3\times-7.1 \approx -1.2\times10^3\text{ J}$$ so the answer is approx. $1.2\times10^3\text{ J}$

Is this correct, or am I missing something?

The next question:

$$m = \dfrac{Q}{c\Delta T}$$

$$m = \dfrac{1.2\times10^3}{4.18\times10^3\times68} \approx 4\text{ mL}$$

This also doesn't seem to be correct. What am I doing wrongly here?

share|cite|improve this question

closed as too localized by Qmechanic, David Z Oct 29 '12 at 2:41

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Hello user, Welcome to Physics.SE. We discourage questions that ask users to solve someone's problems. So, Please be specific on your question. I mean, Consider revision to focus your homework... And then, just start skimming throughout our homework policy on "How to ask homework questions?" – Waffle's Crazy Peanut Oct 28 '12 at 19:45
Hello @CrazyBuddy , I added a sentence that should make my intentions clear. – user14445 Oct 28 '12 at 19:47
Again, you are asking for solving the answers. I think Guys would help only if you're in the wrong path & not when choosing, "WHERE TO GO?" – Waffle's Crazy Peanut Oct 28 '12 at 19:48
Hi user14445, and welcome to Physics Stack Exchange! The original version of your question would definitely be inappropriate here, but after the edits, it's almost fine! The only thing is that you haven't given us any reason to think your answer is incorrect. If you just want someone to check your work for no particular reason, that's not a question for this site, but if you have some reason to expect that your answer is wrong but you can't figure out where your procedure is wrong, then it's okay. (This is kind of a fine line.) – David Z Oct 29 '12 at 2:44
Anyway, if you have some particular reason to believe there is an error in your method (even if that reason is "my answer doesn't agree with the given one, which is ..."), and you've made an honest effort to check your work but can't figure it out, just edit that into your question and flag it for moderator attention so one of us can reopen it. P.S. don't take this personally, in fact you've shown that you are willing to respond to suggestions for improvement on your posts which makes you a likely good contributor IMO ;-) – David Z Oct 29 '12 at 2:46