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Snezhnost [94]
3 years ago
12

A heat engine has a cold reservoir of 250°C and a hot reservoir of 715°C. What is the efficiency of the engine?

Physics
2 answers:
Margarita [4]3 years ago
6 0
If the heat engine is a perfect carnot cycle engine, then
efficiency = (T2-T1)/T2  
where T1=temperature of cold reservoir ( ° K) = 250+273.15 = 523.15
           T2=temperature of hot reservoir ( ° K)  = 715+273.15 = 988.15

Therefore
efficiency = (T2-T1)/T2 = (988.15-523.15)/988.15 = 0.4706 = 47% 

Note: this is the theoretical maximum efficiency of a perfect heat engine.  Real-life engines do not achieve this value.
Illusion [34]3 years ago
5 0

The answer is A. 65.0% for plato users, i got 100% on the test  :)

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Calculate the frequency of visible light having a wavelength of 410 nm
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Definition of AM and Fm Wave? in your own word
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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
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