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I am really stuck on a problem in my textbook:

Water is heated in an open pan where the air pressure is one atmosphere. The water remains a liquid, which expands by a small amount as it is heated. Determine the ratio of the work done by the water to the heat absorbed by the water.

MY ATTEMPT:

We are given that:

$P = 1.013 \cdot 10^5 Pa$

We then have:

$$\frac{W}{Q} = \frac{P \Delta V}{cm \Delta T} = \frac{P \beta V_0 \Delta T}{cm \Delta T} = \frac{P \beta m \Delta T}{ cm \rho \Delta T} = \frac{P \beta}{c \rho} = \frac{1.013 \cdot 10^5 \cdot 207 \cdot 10^{-6}}{4186 \cdot 1} = 5 \cdot 10^{-3}$$

But according to the textbook, the solution should be $4.99 \cdot 10^{-6}$. If anyone can help me by pointing out what I'm doing wrong here, I would be extremely grateful!

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Look, at your units. When something is $10^{-3}$ out, it's defiantly worth checking your units. If you are working in SI units the density of water is $10^3 [kgm^{-3}]$ not $1[kgm^{-3}]$. Imagine 1 metre cubed of water. It's very heavy.

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  • $\begingroup$ Oh, of course! Ugh. Bad mistake on my part. Thanks for pointing it out :). Appreciate it a lot! $\endgroup$
    – user12277
    Sep 22, 2012 at 20:23
  • $\begingroup$ Whether ever you are in your education. Double, or Triple check your units. In an exam they don't sympathise with you, eventually it will become second nature. Where possible always ask yourself, does this sound realistic. $\endgroup$
    – Joshua Siret
    Sep 22, 2012 at 20:25
  • $\begingroup$ Yes, I totally agree with you. And I know perfectly well that the correct density should have been 1000, not 1. It's just that I had been working on a bunch of problems earlier in cm^3, so my mind didn't "switch over" properly when I encoutered this problem. $\endgroup$
    – user12277
    Sep 22, 2012 at 20:40

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