# ratio between work and heat [closed]

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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## closed as too localized by dmckee♦Sep 28 '12 at 20:54

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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.