When heating water, during what temperature range will the temperature cease to change for some time?
1 answer:
Answer: Option (B) is the correct answer.
Explanation:
As we know that the temperature when the vapor pressure of liquid becomes equal to the atmospheric pressure surrounding the liquid. And, during this temperature liquid state of substance changes into vapor state.
But during this process of change in state of substance the temperature will cease to change for some time because unless and until all the liquid molecules do not convert into vapor state the temperature will not rise or change.
As the boiling point of water is so the temperature ceases to change from
to
.
Therefore, we can conclude that when heating water, during to
temperature range the temperature will cease to change for some time.
You might be interested in
Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
where
is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
(c) 5550 m
The maximum playing time of the CD is
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
(d)
The angular acceleration of the CD is given by
where
is the final angular speed (when the CD is scanned at the outermost part)
is the initial angular speed (when the CD is scanned at the innermost part)
is the time elapsed
Substituting into the equation, we find