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dlinn [17]
3 years ago
11

The efficiency of a Stirling cycle depends on the temperatures of the hot and cold isothermal parts of the cycle.If you increase

the upper temperature, keeping the lower temperature the same, does the efficiency increase,decrease, or remain the same?A. increaseB. decreaseC. remain the same
Physics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

A. increase

Explanation:

Stirling cycle having four processes

1.Two processes are constant temperature processes

2.Two processes are constant volume processes

The efficiency of Stirling cycle is same as the efficiency of Carnot cycle.

The efficiency of Stirling cycle given as

\eta=1-\dfrac{T_L}{T_H}

T_L=Lower temperature

T_H=Upper temperature

If we increase then upper temperature while the lower temperature is constant then the efficiency of Stirling cycle will increase because the ratio of lower and upper temperature will decreases.

Therefore answer is

A. increase

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Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
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  • λ = h/(mv)
  • E = (1/2)mv²

Where:

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  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

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           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

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