If the solution is treated as an ideal solution, the extent of freezing
point depression depends only on the solute concentration that can be
estimated by a simple linear relationship with the cryoscopic constant:
ΔTF = KF · m · i
ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF
(solution).
KF, the cryoscopic constant, which is dependent on the properties of the
solvent, not the solute. Note: When conducting experiments, a higher KF
value makes it easier to observe larger drops in the freezing point.
For water, KF = 1.853 K·kg/mol.[1]
m is the molality (mol solute per kg of solvent)
i is the van 't Hoff factor (number of solute particles per mol, e.g. i =
2 for NaCl).
Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity. Its standard units are meters per second per second, or meters per second squared. (this is for number 3)
Answer:
Total distance = 400+700+1200= 2300km
Explanation:
the resultant of d 1st right angle triangle + 1200
= 806.2 + 1200 = 2006.2km
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
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