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Kazeer [188]
3 years ago
8

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

inear expansion coefficient of aluminum is 23.0 x 10^-6 /C°.
Physics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient \alpha=23.0\times10^{-6}/C^{\circ}

We need to calculate the volume expansion coefficient

Using formula of  volume expansion coefficient

\beta=3\alpha

Put the value into the formula

\beta=3\times23.0\times10^{-6}

\beta=0.000069=69\times10^{-6}/C^{\circ}

We need to calculate the volume

V= a^3

V=4^3

V=64\ cm^3

The change temperature of the cube is

\Delta T=T_{f}-T_{i}

Put the value into the formula

\Delta T=67-19 = 48^{\circ}C

We need to calculate the increases volume

Using formula of increases volume

\Delta V=V\beta\Delta T

Put the value into the formula

\Delta V=64\times69\times10^{-6}\times48

\Delta V=0.212\ cm^3

Hence, The volume of an aluminum cube is 0.212 cm³.

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Answer: Thermal Energy

Explanation: I know this answer is correct because I had to look in my notebook for the definition

In my own words: Thermal Energy is the energy coming from heat which means it is generated by the movement of tiny particles with different objects.

Hope this helps! :)

8 0
3 years ago
SI
madam [21]

Answer:

p = mv

Explanation:

  • The momentum of a body is defined as the product of its mass and velocity. Its physical symbol is 'p'.
  • The formula for momentum is given by

                               p = mv

         Where,

                                m -  the mass of the body in kg

                                v - velocity of the body in m/s

  • Therefore, the unit of momentum is expressed as the kg m/s
  • The momentum of a body is always associated with its motion. It is a vector quantity and it is directed in the direction of the velocity vector.
  • If a body is at rest, the momentum associated with the body is zero.
  • The momentum plays a significant role in the kinematics of the body. As similar to the energy conservation law, the total momentum of the body is conserved.
6 0
4 years ago
For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Fi
Ivenika [448]

To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to

T = 0.55s

Therefore the frequency will be the inverse of the period and would be given as

f= \frac{1}{T}

f = \frac{1}{0.55}

f = 1.82s^{-1}

The ground state energy of the pendulum is,

E = \frac{1}{2} hv

E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})

E = 6.03*10^{-34}J

The ground state energy in eV,

E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})

E = 3.8*10^{-15}eV

The energy difference between adjacent energy levels,

\Delta E = hv

\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)

\Delta E = 12.1*10^{-34}J

4 0
3 years ago
Liquid sodium can be used as a heat transfer fluid. Its vapor pressure is 40.0 torr at 633°C and 400.0 torr at 823°C. Calculate
kaheart [24]

Answer:

H vaporization = 100.0788 kJ/mol

Explanation:

Use clausius clapyron's adaptation for the calculation of Hvap as:

ln\frac {P_2}{P_1}=\frac {H_{vap}}{R}(\frac {1}{T_1}-\frac {1}{T_2})

Where,

P₂ and P₁ are the pressure at Temperature T₂ and T₁ respectively.

R is the gas constant.

T₂ = 823°C

T₁ = 633°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature,

T₂ = (823 + 273.15) K = 1096.15 K

T₁ = (633 + 273.15) K = 906.15 K

P₂ = 400.0 torr , P₁ = 40.0 torr

R = 8.314 J/K.mol

Applying in the formula to calculate heat of vaporization as:

ln \frac {400}{40}=\frac {H_{vap}}{R} (\frac {1}{906.15}-\frac {1}{1096.15})

Solving for heat of vaporization, we get:

H vaporization = 100078.823 J/mol

Also, 1 J = 10⁻³ kJ

So,

<u>H vaporization = 100.0788 kJ/mol</u>

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Natasha_Volkova [10]
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 4. Mass of the liquid
 * No. Do not mistake mass for density. As a mental exercise, imagine the buoyant force on a small piece of metal dropped into a swimming pool. Now imagine the buoyant force on that same piece of metal dropped into a lake. In both cases, the buoyant force is the same, yet the lake has a far greater mass of water than the swimming pool. So this is a bad choice.
8 0
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