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Alik [6]
3 years ago
11

1. The current through a light bulb connected across the terminals of a 120 V outlet is 0.50 A. At what rate does the bulb conve

rt electric energy to light? 2. A 12.0 V battery causes a current of 2.0 A to flow through a lamp. What is the power used by the lamp? 3. What current flows through a 100. W light bulb connected to a 120. V outlet? 4. The current through a motor is 210 A. If a battery keeps a 12.0 V potential difference across the motor, what electric energy is delivered to the motor in 10.0 s?
Physics
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

1. 60 W

2. 24 W

3. 0.83 A

4. 25200 J

Explanation:

1. What we are simply asked to look for is Electrical Power. It is the rate electrical energy is being transferred.

It is given as:

P = IV

where I = current and V = voltage.

Therefore, Power is:

P = 0.50 * 120 = 60 W

2. Power, as given in the formula above is:

P = 2.0 * 12

P = 24.0 W

3. According to the formula of Power, current is given as:

I = \frac{P}{V}

Power is 100 W and voltage is 120 V, therefore, current is:

I = \frac{100}{120} \\\\\\I = 0.83 A

4. Recall that power is the time rate of transfer of electrical energy. Mathematically:

P = \frac{E}{t}

where t = time

This means that Electrical energy is:

E = Pt

Recall that Power is:

P = IV

Therefore, Electrical energy is:

E = IVt

E = 210 * 12 * 10\\\\\\E = 25200 J

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Calculate the energy gained by an ice block in the following experiment.
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The heat is exchanged when two different temperature objects come in contact. The energy gained by an ice block is 2.3 Joules.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the object.

A 7g block of ice was added to a coffee cup full of 103.4 grams of water. The water had an initial temperature T₁ = 24.5 C and a final temperature T₂ = 19.2 C after all the ice had melted.

Heat lost by water = Heat gained by ice

Qgain = ms(T₂ -T₁ )

Substituting the value for mass of water m =103.4 g= 0.1034 kg , specific heat of water s = 4.18 kJ/kg and temperature values, we get

Qgain = 0.1034 x 4.18 x (24.5 - 19.2)

Qgain = 2.3 Joules

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2 years ago
Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
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Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

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ΔS total ≥ ΔS al + (-ΔS al) =

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What causes the inner core to be solid?
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A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
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Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

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