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zavuch27 [327]
3 years ago
12

A student made a copper bracelet by hammering a small copper bar into the desired shape. The bracelet has a mass of 30.1 grams a

nd was at a temperature of 21°C in the classroom. After the student wore the bracelet, the bracelet reached a temperature of 33°C. Later, the student removed the bracelet and placed it on a desk at home, where it cooled from 33°C to 19°C. The specific heat capacity of copper is 0.385 J/g•K.
69 Explain, in terms of heat flow, the change in the temperature of the bracelet when the student wore the bracelet. [1]

70 Determine the number of moles of copper in the bracelet. [1]

71 Show a numerical setup for calculating the amount of heat released by the bracelet as it cooled on the desk. [1]

72 Explain, in terms of chemical activity, why copper is a better choice than iron to make the bracelet. [1]
Chemistry
1 answer:
Tcecarenko [31]3 years ago
7 0
69. The student's skin was obviously warmer than the initial temperature of the bracelet. According to the second law of thermodynamics, heat transfers from a higher temperature to a lower temperature hence heat transfers from the student's skin to the bracelet warming up the bracelet. 

70. moles of copper=mass of copper/molar mass of copper
=30.1g/(63.55g/mol)= 0.4736 moles Cu

71. Delta H( heat released) = mCp(deltaT) = 30.1g(0.385 J/gK)(33-19)K = 162.23J

72. Copper is chemically less active than iron. 
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Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

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E in J/mol of ejected electrons

Ecalculated × Avogadros constant

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c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

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