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3 years ago

Does potential energy increase with temperature?

2 answers:

6 0

-- The potential energy of a 12-lb bowling ball up on the shelf
doesn't have anything to do with the temperature of the ball or
the shelf.

-- The potential energy of a jar full of gas does depend on the
temperature of the gas. The warmer it is, the greater its pressure
is, and the more work it can do if you let it out through a little hole
in the jar. If it gets hot enough, it'll have enough potential energy
to blow the jar to smithereens.

Wewaii [24]3 years ago

3 0

When a solid melts and becomes a liquid, we say it changes phase from a solid to a liquid. In this change, the bonding between the atoms or molecules changes. You have to "break" some bonds to go from a solid to a liquid. This requires energy. The liquid is a "higher" potential energy state than the solid, even at the same temperature. (It is slightly more complicated than this, but this is good enough for this class.) To convert 1kg of solid water at 0oC (273K) to liquid water at 0oC (273K) requires about 330,000J of energy. Note that the temperature of the liquid is the same as the solid’s, i.e. you added heat without changing the temperature, instead the phase changed. The heat added went into "breaking" bonds and increasing its potential energy, not into increasing the average translational KE of the molecules. (It is slightly more complicated than this, but this explanation is good enough for this class.) If you go the other way, and convert 1kg of liquid water at 0oC (273K) to ice at 0oC (273K) releases 330,000J of heat. This heat comes from the energy given off when bonds form, i.e. it goes to a state of lower potential energy. The same type of thing occurs when a liquid changes to a gas. Then more bonds are broken as the molecules move apart, and it requires energy to break the bonds and move to a higher potential energy. To convert 1kg of liquid water at 100oC (373K) to 1kg of water vapor at 100oC (373K) requires 2,260,000J (almost 2.3 million Joules) of energy. That is, the water absorbs energy to change from a liquid to a gas.. If 1kg of water vapor conde

You might be interested in

What energy change is associated with the reaction to obtain one mole of H2 from one mole of water vapor? The balanced equation

Yakvenalex [24]

Answer:

ΔH = 249 kJ/mol

Explanation:

The balanced reaction is:

2H₂O(g) → 2H₂(g) + O₂(g) (1)

To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:

H₂O(g) → H₂(g) + 1/2O₂(g) (2)

The enthalpy of the reaction (2) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$

Where $\Delta H_{r}$ : is the bond enthalpy of reactants and $\Delta H_{p}$ : is the bond enthalpy of products.

For the reactants we have the next bond energies:

2 x (H-O) = 2 x (467)

And the bond energies for the products are:

H-H + (1/2) (O=O) = 436 + (1/2)(498)

So, the enthalpy of the reaction (2) is:

$\Delta H = 2 \cdot 467 kJ/mol - 436 kJ/mol - \frac{1}{2} \cdot 498 kJ/mol = 249 kJ/mol$

I hope it helps you!

3 0

3 years ago

A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3

Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0

3 years ago

Blood vessels that carry blood away from the heart are called

White raven [17]

Answer:

The blood vessels dat carry blood away from the heart are non as arteries, while those dat carry blood back to the heart are veins.

Explanation:

The ARTERIES are major blood vessels connected to your heart.

8 0

3 years ago

What is the measure of how much a material resists the formation of an electric field?

shusha [124]

The answer is c capacitance

5 0

4 years ago

A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first

Dominik [7]

Answer:

a) $V=14.01 m/s$

b) $V=8.86 \, m/s$

c) $t = 2.33s$

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

$E_m = E_k +E_p$

That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this $E_m$ mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

$E_k = \frac{1}{2} m \, V^2$

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

$E_p= m\, g\, h$

where g is the acceleration due to gravity ( $g=9.81 \, m/s^2$ ) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which $h=0$ , a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

$E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m$

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set $h=0$

$E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2$

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

$E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\$

And so we have found the velocity of the ball as it hits the floor.

$V = \sqrt[]{2g\cdot 10m}=14.01\, m/s$

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

$V = \sqrt[]{2g\cdot h}$

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

$V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s$

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

$V_{avg}\cdot t=h\\t=h/V_{avg}$