Answer:
14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.
Explanation:
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.
During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.
So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.
In this case, being:
- L= 84
and replacing in the expression Q = m*L you get:
Q=172 g*84
Q=14,448 J
<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>
Answer:try to make a good pic please I have ideas about the topic
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
Answer:
189.2 KJ
Explanation:
Data Given
wavelength of the light = 632.8 nm
Convert nm to m
1 nm = 1 x 10⁻⁹
632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m
Energy of 1 mole of photon = ?
Solution
Formula used
E = hc/λ
where
E = energy of photon
h = Planck's Constant
Planck's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light
speed of light = 3 × 10⁸ ms⁻¹
λ = wavelength of light
Put values in above equation
E = hc/λ
E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)
E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)
E = 3.141 x 10⁻¹⁹J
3.141 x 10⁻¹⁹J is energy for one photon
Now we have to find energy of 1 mole of photon
As we know that
1 mole consists of 6.022 x10²³ numbers of photons
So,
Energy for one mole photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³
Energy for one mole photons = 1.89 x 10⁵ J
Now convert J to KJ
1000 J = 1 KJ
1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ
So,
energy of one mole of photons = 189.2 KJ
Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)
Thus the heat of vaporization is 41094 Joules
