Answer to Question #122499 in Molecular Physics | Thermodynamics for Thandeka

Question

Water is heated in an open pan where the air pressure is 1.00 atmosphere. The water
remains a liquid, which expands by a small amount as it is heated. Calculate the ratio
of the work done by the water to the heat absorbed by the water.

Answer 1

Given : $P = 1 atm = 1.00\times 10^5Pa.$

$W$ = Work done

$Q=$ Heat absorbed

$P =$ Air Pressure

$c =$ specific heat of water s 1 calorie/gram $°C = 4.186 \frac {joule}{gram °C}$

$\beta=$ compressibility of water $=207\times 10^{-6}$

$\rho =$ Density of water  1000 $\frac {kg}{m^3}$

$\frac{W}{Q} = \frac{P \Delta V}{cm \Delta T} = \frac{P \beta V_0 \Delta T}{cm \Delta T} = \frac{P \beta m \Delta T}{ cm \rho \Delta T} = \frac{P \beta}{c \rho} = \frac{1.00 \cdot 10^5 \cdot 207 \cdot 10^{-6}}{4186 \cdot 1} = 4.94 \cdot 10^{-3}$