Question

one gram of water becomes 1671cm³ of steam when boiled at a pressure of one atmospheric. if the specific latent heat of vaporization on the water is 2.26 j/kg Calculate (a) the external work done (b) the increase in external energy.

Answer 1

Given

m = 1g

Lvap = 2260 j/g

p = 1.01 x 10^5 Pa = 1 Atm

T = 100 C = 373 K

Vsteam = 1671 cm3

Vwater = 1 cm3

ΔV = (Vsteam - Vwater) = 1671 - 1 = 1670 cm3

Procedure

a) External work done by the system

Q = mL

$\begin{gathered} Q=1g\cdot2260\frac{J}{g} \\ Q=2260J \end{gathered}$

W = pΔV

$\begin{gathered} W=1.01\times10^5\cdot1670\times10^{-6} \\ W=169\text{ J} \end{gathered}$

External word done by the system is 169 J

b) increase in internal energy

ΔU = Q - W

$\begin{gathered} \Delta U=2260J-169J \\ \Delta U=2091J \end{gathered}$

Increase in the internal energy is 2091 J