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Nutka1998 [239]
2 years ago
5

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1

000g=1kg)
b. How many grams of carbon are present if it requires 37.5 kJoules of energy to heat a sample from 30.0 to 55.0C ? (specific heat of carbon is 0.71 J/g C)

c. If 480.0 Joules is applies to a 50.0g sample of Hg in a thermometer that reads 25.0C, what will be the final temperature of the sample? (SH mercury is 0.14)

d. A 25.0kg sample of an unknown metal X requires 875kJ of energy to heat it from 55.0 to 125.0C. What is the specific heat of the unknown metal?


5.690x103 miles to meters.

900.0kg to grams

Please also give sig figs to all answers.

Answer fast please.
Chemistry
2 answers:
DochEvi [55]2 years ago
7 0

a.

A substance's specific heat tells you how much heat is required to increase the temperature of 1 g of that substance by 1°C.

The equation that establishes a relationshop between heat and change in temperature is

q = m • c • ∆T, where

q - heat absorbed

c - the specific heat of the substance, in your case of water

ΔT - the change in temperature, defined as the difference between the final temperature and the initial temperature

so:

q = 1.00 g • 4.18 J/g×°C • (75.0 - 38.5)°C

q = 152,57 J

just apply this formula for all exercises

Marta_Voda [28]2 years ago
5 0

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1000g=1kg)

Answer:

q = 534,506 Joules

q = 534.506 KiloJoules

534.506 KiloJoules or 534,506 Joules energy required to raise 3.50kg of water from 38.5 to 75.0C

<h3>What is sp.heat ?</h3>

Sp. Heat is the heat required to increase the temperature of the 1 mass of a given substance by a  1^{0} C temperature.

The formula of specific heat Cp = \frac{q}{m (∆T)}

Where,

q = energy of substance (Joules / KiloJoules),

Cp = Specific heat capacity of the substance (J/Kg.C),

m = mass of the substance

∆T = Change in temp.

Explanation:

Given data from que: -

Mss of the water (m) = 3.50 Kg = 3.50 × 1000 gm = 3500 gm

specific heat capacity of water (Cp) = 4.184J/g C

Change in temp (∆T) = 75.0 - 38.5 = 36.5

now, put all above given data in formula

we get

Cp = \frac{q}{m (∆T)}

q = Cp ×m×∆T

q = 4.184×3500×36.5

q = 534,506 Joules

q = 534.506 KJ

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In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
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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

Hope this Helps!!!

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