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Serjik [45]
2 years ago
5

Suppose an oven's radiation wavelength is 0.125 m. a container with 350.00 g of water was placed in the oven, and the temperatur

e of the water rose from 20.0°c to 80.0°c. how many photons of this microwave radiation were required to heat the water? (assume that all the energy from the radiation was used to raise the temperature of the water.) the specific heat of water is 4.18 j/g°c.
Physics
1 answer:
ivann1987 [24]2 years ago
8 0
The heat (energy) needed to raise the temperature of the water is given by
Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J

The wavelength of the radiation of the oven is \lambda=0.125 m, so the energy of a single photon of this radiation is
E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34}J) \frac{3\cdot 10^8 m/s}{0.125 m}=1.6 \cdot 10^{-24} J

So, the number of photons required to heat the water is the total energy absorbed by the water divided by the energy of a single photon:
N= \frac{Q}{E}= \frac{87780 J}{1.6\cdot 10^{-24}J}=  5.5 \cdot 10^{28} photons
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d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
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2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
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