How many grams of water can be cooled from 42 ∘c to 20 ∘c by the evaporation of 51 g of water? (the heat of vaporization of wate
1 answer:
Let us first calculate heat obtained by the evaporation of 51 g of water.
Given, heat of vaporization of water = 2.4 kJ/ g
∴ Heat obtained by evaporation of 51 g of water = 2.4 × 51 = 122.4 kJ
This is the heat energy available that can be used to cool water from 42°C to 20°C.
Specific heat of water is given by,
Here,
C is the specific heat of water = 4.18 J/gK
Q is the amount of heat = 122400 J
m is the mass of the water that can be cooled.
dt is the change in temperature= 42°C ₋ 20°C = 22°C ( The numerical value will be the same if Kelvin unit is used.)
Substituting the values we get,
m = 1331 g
1331 grams of water can be cooled from 42°C to 20°C by evaporation of 51 g of water.
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Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
=
= 4.189 rad/s
Angular acceleration α =
=
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
=
= 25.97 m/s
So centripetal acceleration.
a =
=
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.
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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?
Data:
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
Answer:
The displacement of the spring = 0.8 m
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