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telo118 [61]
3 years ago
11

A pie is cooked in an oven at 200 °C. The aluminium film that ckvered the pie can be touched soon after it is removed while the

pie is still dangerously hot. Explain this?
Physics
1 answer:
GREYUIT [131]3 years ago
8 0

Answer and Explanation:

The aluminum is more productive in the absorption and heat transfer to other particles. It instantly converts heat absorbed from the environment into the atmosphere when removed from the oven, enabling us to operate with it faster than the pie that takes much longer to convert heat to the environment.

So this is the reason for pie to be the dangerously hot

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how much thermal energy is absorbed by a copper water pipe with a mass of 2.3kg when its temperature is raised from 20.0 to 80.0
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Explanation:

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Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of
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a) 1.59(10)^{-19} J

b) 2.34(10)^{12} electrons

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.  

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy  E:

E=\frac{hc}{\lambda} (1)  

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the kinetic energy K of the photoelectron:  

E=\Phi+K (2)  

Where \Phi=6.94(10)^{-19} J is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a)  Maximum possible kinetic energy of the emitted electrons (K)</h3>

From (1) we can know the energy of one photon of 233 nm light:

E=\frac{hc}{\lambda}

Where:

h=6.63(10)^{-34}J.s is the Planck constant  

\lambda=233 (10)^{-9} m is the wavelength

c=3 (10)^{8} m/s is the speed of light

E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s} (3)

E=8.53(10)^{-19} J (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

8.53(10)^{-19} J=6.94(10)^{-19} J+K (5)  

Finding K:

K=1.59(10)^{-19} J (5)  This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is 2 \mu J=2(10)^{-6} J</h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

\frac{E_{burst}}{E}

Where E_{burst}=2(10)^{-6} J is the energy of the burst of light

Hence:

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