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Ch.5 - Periodicity & Electronic Structure of Atoms

Chapter 5, Problem 131

Microwave ovens work by irradiating food with microwave radiation, which is absorbed and converted into heat. Assum-ing that radiation with l = 15.0 cm is used, that all the energy is converted to heat, and that 4.184 J is needed to raise the temperature of 1.00 g of water by 1.00 °C, how many photons are necessary to raise the temperature of a 350 mL cup of water from 20 °C to 95 °C?

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hi everyone for this problem. It reads microwave ovens operate by exposing food to microwave radiation which is absorbed and turned into heat. How many photons are required to raise the temperature of a millimeter cup of water from 23.5 degrees Celsius to 97.2 degrees Celsius. If radiation with a length equal to 10 centimeters is used, all energy is converted to heat and 4.184 jewels is required to raise the temperature of one g of water by one degree Celsius. So the question that we want to answer here is the amount of photons required to raise the temperature. Okay, in order for us to calculate this, we first need to calculate the amount of heat gained or lost by the sample. And we can calculate this by using the equation Q. Is equal to M C delta T. Okay. Where Q represents heat, M. Is the mass of the sample. C. Is the specific heat and delta T. Is the temperature change. Alright, so for our mass, our mass is going to equal the mass of water. And what we can do here is we can use the density of water Which is one g per mil, a leader of water to convert the volume to mass. So let's go ahead and do that. Our volume is mL. And we want to go from volume to mass. So let's go ahead and plug in that density. So our conversion is the density of water is one g per millimeter. And we want to go from male leaders to leaders in one leader. There is 1000 milliliters. So that means that our mass when we convert it from volume two g is going to equal 0.450 g. So let's go ahead and plug in what we know into our equation. So our heat is going to equal our mass which is 0.450 g times the specific heat which is in the problem we're told it's 4.184 jewels per gram degree Celsius Times are changing temperature. So our temperature is changing from 23.5°C to 97.2°C. So we need to take the difference. So we're going to take 97.2°C. And we're going to subtract 23.5°C. So let's do this calculation. And when we do this calculation we're going to get 138.76 - 6 jules. Okay, so this is the heat now that we know the heat we can use the equation for or that shows the relationship between energy of a photon and wavelength of light. And that equation is energy equals Planck's constant times the speed of light over wavelength. Alright, so let's go ahead and write down The units here. So H represents Plank's constant and this value is 6.626 times 10 to the negative 34 jewels per second. Okay. Or Jewels Times 2nd. C represents speed of light and this is 3.00 times 10 to the eight m/s and lambda equals wavelength. Okay. And as you can see our speed of light is in meters per second. So that means our wavelength needs to be in meters. And the problem we're told the wavelength is 10 centimeters. So we need to go from centimeters to meters in one centimeter. There is 10 to the negative two m. So our units for centimeters cancel. And so our wavelength and meters is going to equal 0. m. And we just calculated energy. Okay, the energy excuse me. No, we're calculating energy. So let's go ahead and plug in our values. And what, what the value we're going to get is the value for energy. So energy is equal to Planck's constant. 6.626 times 10 to the negative 34 jules times seconds times our speed of light which is 3.0 times 10 to the eight m per second. And this is all over wavelength. Okay, so this is over 0.1 m. So when we calculate this, the value that we're going to get for energy is going to equal 1.98. Okay, so this is going to equal 1. Times 10 to the negative jewels per photon. So this is the energy and jewels per photon But we want to know how many photons, so we just want to know the number of photons. Alright, so for us to get the number of photons, what we're going to do is our number of photons is going to equal our heat that we calculated, divided by our energy that we calculated. Alright, so let's go ahead and plug those values in. The heat that we calculated is 138.76 to 36 jewels. And the energy that we just calculated is 1.987, 8 times 10 to the -20 for jewels per photon. Okay, so as you can see here are jewels cancel and we're left with the unit of photon. So let's go ahead and do this calculation. And when we do this calculation we're going to get 6.9807 times to the 25 photons as our final answer. And this is going to be it for this problem. This is how many photons are required to raise the temperature. Okay, so that is it for this problem. I hope this was helpful
Related Practice
Textbook Question

Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except that the angular-momentum quantum number l can have integral values of 0, 1, 2...n + 1 (instead of 0, 1, 2..., n - 1). (a) How many elements would be in the first two rows of the periodic table in this universe?

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Textbook Question

Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except that the angular-momentum quantum number l can have integral values of 0, 1, 2...n + 1 (instead of 0, 1, 2..., n - 1). (c) Draw an orbital-filling diagram for the element with atomic number 12.

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Textbook Question
One watt (W) is equal to 1 J/s. Assuming that 5.0% of the energy output of a 75 W light bulb is visible light and that the average wavelength of the light is 550 nm, how many photons are emitted by the light bulb each second?
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Textbook Question
The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy can be measured by a technique called photoelectron spectroscopy, in which light of wavelength l is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron (Ek) is measured by determining its veloc-ity, v (Ek = mv2/2), and Ei is then calculated using the conservation of energy principle. That is, the energy of the incident light equals Ei plus Ek. What is the ionization energy of selenium atoms in kilojoules per mole if light with l = 48.2 nm produces electrons with a velocity of 2.371 * 106 m/s? The mass, m, of an electron is 9.109 * 10-31 kg.
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Textbook Question
X rays with a wavelength of 1.54 * 10-10 m are produced when a copper metal target is bombarded with high-energy electrons that have been accelerated by a voltage difference of 30,000 V. The kinetic energy of the electrons equals the product of the voltage difference and the electronic charge in coulombs, where 1 volt-coulomb = 1 J. (a) What is the kinetic energy in joules and the de Broglie wavelength in meters of an electron that has been accel-erated by a voltage difference of 30,000 V?
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Textbook Question
In the Bohr model of atomic structure, electrons are constrained to orbit a nucleus at specific distances, given by the equation

where r is the radius of the orbit, Z is the charge on the nucleus, a0 is the Bohr radius and has a value of 5.292 * 10-11 m, and n is a positive integer (n = 1, 2, 3...) like a principal quantum number. Furthermore, Bohr concluded that the energy level E of an electron in a given orbit is

where e is the charge on an electron. Derive an equation that will let you calculate the difference ∆E between any two energy levels. What relation does your equation have to the Balmer–Rydberg equation?
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