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24.6 ℃
<h3>Explanation</h3>Hydrochloric acid and sodium hydroxide reacts by the following equation:
which is equivalent to
The question states that the second equation has an enthalpy, or "heat", of neutralization of . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce
or
of energy.
500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release of energy.
Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.
The question has given the heat capacity of the calorimeter directly.
The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.
The calorimeter contains 1.00 liters or of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of
.
The solution has a specific heat of . The solution thus have a heat capacity of
. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.
The calorimeter-solution system thus has a heat capacity of , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy.
are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.
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
h = Planck's constant = 6.63 × 10⁻³⁴ J.s
E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J
E in J/mol of ejected electrons
Ecalculated × Avogadros constant
= 2.92 × 10⁻¹⁹ × 6.023 × 10²³
= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons
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
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