Question

Water in an electric teakettle is boiling. The power absorbed by the water is 0.90 kW. Assuming that the pressure of vapor in the
kettle equals atmospheric pressure, determine the speed of effusion of vapor from the kettle's spout if the spout has a cross-
sectional area of 1.60 cm2. Model the steam as an ideal gas.​

Answer 1

Answer:

v = 4.233  m/s

Explanation:

By applying the rate of boiling from $Q= mL_v$;

the rate of the boiling can be described as:

$\mathcal{P} = \dfrac{Q}{\Delta t} \\ \\ \mathcal{P} = \dfrac{mL_v}{\Delta t}$

The mode of the steam (water vapor) as an ideal gas can be illustrated by formula:

$P_oV_o = nRT$ --- (1)

where;

n = number of moles;

$n = \dfrac{mass (m)}{Molar mass (M)}$

Then; equation (1) can be rewritten as:

$P_oV_o = (\dfrac{m}{M}) RT \\ \\ \dfrac{P_oV}{\Delta T} = \dfrac{m}{\Delta t} ( \dfrac{RT}{M})$

∴

$\dfrac{m}{\Delta t} = \dfrac{\mathcal{P}}{L_v}$

Then:

$P_o \times A \times v= \dfrac{\mathcal{P}}{L_v}\Big ( \dfrac{RT}{M }\Big)$

making (v) the subject of the formula:

$v= \Big ( \dfrac{\mathcal{P} RT}{M\times L_v \times P_o \times A }\Big)$

Given that:

$\mathcal{P}$ = 0.90  kW = 900 W

R(rate constant) = 8.314 J/mol.K

Temperature at 100° C = 373K

For water vapor:

molar mass= 18.015 g/mol ≅ 0.0180 kg/mol

Latent heat of vaporisation $L_v$ = 2.26 × 10⁶ J/kg

Atmospheric pressure $P_o = 1.013 \times 10^6 \ N/m^2$

Cross sectional area A =1.60 cm² = 1.60 × 10⁻⁴ m²

$v= \Big ( \dfrac{900 W (8.314 \ J/mol.K)(373)}{0.0180 \ kg/mol) (2.26 \times 10^6 \ J/kg) (1.013 \times 10^5 \ N/m^2)(1.60 \times 10^{-4} \ m^2)}\Big)$

v = 4.233  m/s