How much heat is released when 432 g of water cools down from 71'c to 18'c?
1 answer:
The heat released by the water when it cools down by a temperature difference 
is
where
m=432 g is the mass of the water
is the specific heat capacity of water
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.
where
m=432 g is the mass of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.
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To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,
Here,
= Wavelength
d= Diameter of aperture
= Angular resolution or diffraction angle
Our values are given as,
The frequency of the sound is
The speed of the sound is
The wavelength of the sound is
Here,
v = Velocity of the wave
f = Frequency
Replacing,
The diffraction condition is then,
Replacing,
d = 0.24 m
Therefore the diameter should be 0.24m
Answer:
The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is
Explanation:
Here we have
Initial temperature of air T₁ = 20 °C = 293.15 K
Final temperature of air T₁ = 120 °C = 393.15 K
Initial pressure P₁ = 1 atm = 101325 Pa
Final pressure P₂ = Required
Area = A
Therefore we have for the pressure cooker, the volume is constant that is does not change
By Chales law
P₁/T₁ = P₂/T₂
P₂ = T₂×P₁/T₁ = 393.15 K× (101325 Pa/293.15 K) = 135,889.22 Pa
∴ P₂ = 135.88922 KPa = 135.9 kPa
Where Force = we have
Force = .