Problem 7.31- Energy Per Photon & Amount of Energy Needed to

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Hannah Pablo 2H
Posts: 10
Joined: Fri Sep 26, 2014 2:02 pm

Problem 7.31- Energy Per Photon & Amount of Energy Needed to

Postby Hannah Pablo 2H » Sun Jan 11, 2015 12:18 pm

Problem:
In a microwave oven, radiation is absorbed by water in the food and the food is heated. How many photons of wavelength 4.50 mm are required to heat 350. g of water from 25.0°C to 100.0°C, assuming all the energy is used to raise the temperature?

How do I figure out the energy needed to heat water?

Erika Monasch 4I
Posts: 5
Joined: Fri Sep 26, 2014 2:02 pm

Re: Problem 7.31- Energy Per Photon & Amount of Energy Neede

Postby Erika Monasch 4I » Sun Jan 11, 2015 1:00 pm

You would use the equation q=mCdeltaT to find the heat required to heat the water. For the equation, C is 4.184 J/(degrees C*g), which is the specific heat capacity of liquid water. Did that help?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Problem 7.31- Energy Per Photon & Amount of Energy Neede

Postby Niharika Reddy 1D » Sun Jan 11, 2015 1:02 pm

The energy required to heat the water can be found by using the equation q=mCΔT. The mass is 350.g, the specific heat of liquid water is 4.184 J/(g°C), and ΔT is 100.0°C-25.0°C=75.0°C. Multiplying these values will give the energy required to heat the water: 1.098x10^5 J.

I have attached a picture of my work. I carried multiple decimal places and just accounted for significant figures at the very end for accuracy. Hope this helps!
Attachments
image.jpg
7.31

lilyhui3I
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Joined: Tue Nov 25, 2014 3:00 am

Re: Problem 7.31- Energy Per Photon & Amount of Energy Neede

Postby lilyhui3I » Mon Jan 26, 2015 5:36 pm

Where did the equation E=(hc)/ λ come from. I assumed it to be KE=hv-ϕ where KE=0 and v=c/λ. Is that correct?

Maria Davila 2I
Posts: 12
Joined: Fri Sep 26, 2014 2:02 pm

Re: Problem 7.31- Energy Per Photon & Amount of Energy Neede

Postby Maria Davila 2I » Mon Jan 26, 2015 10:26 pm

When you are finding the energy of a photon you make use of two equations:
and
you then substitute v (in the second equation) for
which another way of writing equation 1. Now you can use the wavelength which you will have to turn into meters to cancel out the meter unit in 'c' - speed of light constant.

Hope this helps in understanding why we use these two equations.


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