Answer:
v = 4.233 m/s
Explanation:
By applying the rate of boiling from [tex]Q= mL_v[/tex];
the rate of the boiling can be described as:
[tex]\mathcal{P} = \dfrac{Q}{\Delta t} \\ \\ \mathcal{P} = \dfrac{mL_v}{\Delta t}[/tex]
The mode of the steam (water vapor) as an ideal gas can be illustrated by formula:
[tex]P_oV_o = nRT[/tex] --- (1)
where;
n = number of moles;
[tex]n = \dfrac{mass (m)}{Molar mass (M)}[/tex]
Then; equation (1) can be rewritten as:
[tex]P_oV_o = (\dfrac{m}{M}) RT \\ \\ \dfrac{P_oV}{\Delta T} = \dfrac{m}{\Delta t} ( \dfrac{RT}{M})[/tex]
∴
[tex]\dfrac{m}{\Delta t} = \dfrac{\mathcal{P}}{L_v}[/tex]
Then:
[tex]P_o \times A \times v= \dfrac{\mathcal{P}}{L_v}\Big ( \dfrac{RT}{M }\Big)[/tex]
making (v) the subject of the formula:
[tex]v= \Big ( \dfrac{\mathcal{P} RT}{M\times L_v \times P_o \times A }\Big)[/tex]
Given that:
[tex]\mathcal{P}[/tex] = 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 [tex]L_v[/tex] = 2.26 × 10⁶ J/kg
Atmospheric pressure [tex]P_o = 1.013 \times 10^6 \ N/m^2[/tex]
Cross sectional area A =1.60 cm² = 1.60 × 10⁻⁴ m²
[tex]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)[/tex]
v = 4.233 m/s
