Question

$4 \text{ kg}$ of water is heated from $4^\circ \text{C}$ to $20^\circ \text{C}$ at constant pressure $10^5 \text{ Pa}$ so that density changes from $1000 \text{ kg/m}^3$ to $998 \text{ kg/m}^3$. Then find $\Delta U$ (in $\text{Joules}$) given $C_v$ of $\text{H}_2\text{O}= 4.2 \text{ Joule/gm.K}$ :

• A. $268799.2 \text{ Joule}$
• B. $368900 \text{ Joule}$
• C. $168400 \text{ Joule}$
• D. $578876.8 \text{ Joule}$

Hint:

For liquid heating problems, the work done due to volume change is very small but must be accounted for using $w=-P\Delta V$. Ensure all units are consistent (Joules and Pascals/m$^3$).

Answer 1

The Correct Option is A

Calculate heat absorbed ($q$): $q = m c \Delta T$. ($m=4000 \text{ g}, c \approx 4.2 \text{ J/gK}, \Delta T = 16 \text{ K}$).
$q = 4000 \times 4.2 \times 16 = 268800 \text{ J}$.
Calculate work done ($w$): $w = -P \Delta V = -P (\frac{m}{\rho_2} - \frac{m}{\rho_1})$.
$w = -10^5 \text{ Pa} \times \left( \frac{4 \text{ kg}}{998 \text{ kg/m}^3} - \frac{4 \text{ kg}}{1000 \text{ kg/m}^3} \right)$.
$w = -10^5 \times 4 \times \left( \frac{1000 - 998}{998 \times 1000} \right) \text{ m}^3 \approx -10^5 \times 4 \times (2 \times 10^{-6}) \text{ m}^3$.
$w \approx -0.8 \text{ J}$.
Calculate change in internal energy ($\Delta U$): $\Delta U = q + w$.
$\Delta U = 268800 \text{ J} - 0.8 \text{ J} = 268799.2 \text{ Joule}$.