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2 years ago

An 8.38 gram sample of water lost 870 J of heat. What is the change in temperature

Physics

1 answer:

Virty [35]2 years ago

6 0

Answer:

Find the mass of a sample of water if its temperature dropped 24.8 degrees Celsius when it lost 870 J of. heat. (specific heat of water is 4.18 J/g x degrees Celsius) 8.38 grams. How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of.

Explanation:

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2 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

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Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

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3 years ago

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

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Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

$E=\frac{hc}{\lambda}$

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

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2 years ago

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