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3 years ago

How would you describe the appearance of the substance after the phase change?

Physics

1 answer:

Leona [35]3 years ago

4 0

Answer:

Substances can change phase—often because of a temperature change. At low temperatures, most substances are solid; as the temperature increases, they become liquid; at higher temperatures still, they become gaseous. The process of a solid becoming a liquid is called melting.

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Which statement best describes the law of conservation of energy?

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NEED HELP ASAP <br> 2. What is an example of a question that cannot be solved using science?

Goryan [66]

Answer:

How did life begin on earth?

Explanation:

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2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially

Naddika [18.5K]

Answer:

E_particle = 1,129 10⁻²⁰ J / particle

T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

E_a = 2 5000 = 10000 J

Gas b

E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

E_total = E_a + E_b

E_total = 10000 + 24000 = 34000

The total mass is

M = m_a + m_b

M = 2 +3 = 5

The average energy among the entire mass is

E_averge = E_total / M

E_averago = 34000/5

E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

E_particle = 1,129 10⁻²⁰ J / particle

For find the temperature let's use equation

E = kT

T = E / k

T = 1,129 10⁻²⁰ / 1,381 10⁻²³

T = 8.175 102 K

T= 817.5 K

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3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

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3 years ago