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!