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InterviewSolution
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What flows through a garden hose at a volume flow rate dV/dt of 450cm^(3)//s. What |
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Answer» Solution :The current I of negative charge is due to the electrons in the water molecules moving through the hose. The current is the RATE at which that negative charge passes through any plane that cuts COMPLETELY across the hose. Calculations: Calculations: We can write the current in terms of the number of molecules that pass through such a plane per second as i=(charge per electron) (electrons per molecule) (molecules per second) or `i=(e) (10) (d N)/(dt)` We SUBSTITUTE 10 electrons per molecule because a water `(H_2O)` molecule contains 8 electrons in the SINGLE oxygen atom and 1 electron in each of the two hydrogen atoms. We can express the rate dN/dt in terms of the given volume flow rate dV/dt by first writing (molecule per second)=(molecules per mole) (moles per unit mass) `xx ("mass per unit volume") ("volume per second")` "Molecules per mole" is Avogadro.s number `N_A.` "Moles per unit mass" is the inverse of the mass per mole, which is the molar mass M of water. "Mass per unit volume" is the (mass) density `p_("mass")` of water. The volume per second is the volume flow rate dV/dt. Thus, we have `(d N)/(dt)=N_(A) (1/M)p_("mass") ((d V)/(dt))=(N_(A) p_("mass"))/(M) (d V)/(dt)` SUBSTITUTING this into the equation for i, we find `i=10eN_(A) M^(-1) p_("mass") (d V)/(dt)` We know that Avogadro.s number `N_A.` is `6.02 xx 10^(23)` molecules/mol, or `6.02 xx 10^(23)" mol"^(-1),` and from Table 15-1 we know that the density of water `p_("mass")` pris under normal conditions is 1000 kg/m`""^(3)`. We can get the molar mass of water from the molar masses listed in Appendix F (in grams per mole): We add the molar mass of oxygen (16 g/mol) to twice the molar mass of hydrogen (1 g/mol), obtaining 18 g/mol = 0.018 kg/mol. So, the current of negative charge due to the electrons in the water is `i=(10) (1.6 xx 10^(-19C)) (6.02 xx 10^(23) "mol"^(-1)) xx (0.018 kg//mol)^(-1) (1000 kg//m^(3)) (450 xx 10^(-6) m^(3)//s)` `=2.41 xx 10^(7) C//s=2.41 xx 10^(7)A` =24.1MA. This current of negative charge is exactly compensated by a current of positive charge associated with the nuclei of the three atoms that make up the water molecule. Thus, there is no net flow of charge through the hose. |
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