How much heat is needed to vaporize 33.3 grams of ethyl alcohol at its boiling point of 78.0°C? The latent heat of vaporization
2 answers:
Answer:
Q = 28538.1 J
Explanation:
As we know that heat require to vaporize the ethyl alcohol is given as
Q = mL
here we know that
m = mass = 33.3 gram
L = latent heat of fusion = 857 J/g
now from above formula
so heat required in three significant figures is given as
so heat required to vaporize ethyl alcohol is
You might be interested in
1. -23.2 J
The gravitational potential energy of the ball is given by
where
m = 1.1 kg is the mass of the ball
g = 9.8 m/s^2 is the acceleration of gravity
h is the height of the ball, relative to the reference point chosen
In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is
h = -2.16 m
And the gravitational potential energy is
2. 41.1 J
Again, the gravitational potential energy of the ball is given by
In this part of the problem, the reference point is the floor.
The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:
h = 5.97 m - 2.16 m = 3.81 m
And so the gravitational potential energy of the ball relative to the floor is
3. 0 J
As before, the gravitational potential energy of the ball is given by
Here the reference point is a point at the same elevation of the ball.
This means that the heigth of the ball relative to that point is zero:
h = 0 m
And so the gravitational potential energy is