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Dima020 [189]
3 years ago
14

How many electrons pass through a standard 60 watt light bulb in one second?

Physics
1 answer:
timofeeve [1]3 years ago
6 0
A standard 60 W light bulb has a voltage of 130 volts. So, we use this conversion, the Faraday's constant which is equal to approximately 96,500 Coulombs per mole electron, and the Avogadro's number equal to 6.022×10²³ particles/mole . The solution is as follows:

W = Energy/time
60 W = x J/1 s
x = 60 J = 60 C·V

(60 C·V)*(1/130 V)*(1 mole e/96,500 C)*(6.022×10²³ electrons/mole electron)
= 2.88×10¹⁸ electrons
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E = (1/2)Iω²  

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ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0
3 years ago
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