Pass through it easily is called a Metals such as iron, steel, copper, and aluminum are good conductors.
Answer:
The power output of this engine is 
The the maximum (Carnot) efficiency is 
The actual efficiency of this engine is 
Explanation:
From the question we are told that
The temperature of the hot reservoir is 
The temperature of the cold reservoir is 
The energy absorbed from the hot reservoir is 
The energy exhausts into cold reservoir is 
The power output is mathematically represented as

Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as

substituting values

So


The Carnot efficiency is mathematically represented as



The actual efficiency is mathematically represented as

substituting values


Answer:
(a) W=1.20×10⁴J
(b) U= -5.46×10⁴J
(c) Q= -4.26×10⁴J
Explanation:
Given that student does 1.20×10⁴J work
(a) W=1.20×10⁴J
Work done by student,so positive sign
During the process, his internal energy decreases by 5.46×10⁴J.
(b) U= -5.46×10⁴J.
As the Energy decreases therefore negative sign
For (c) Q
We know the formula

Answer:
1.F: About 6*10^14 Hz
2.E: About 4*10^ -19 J
Explanation:
Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz). By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)
=6.05*10^14 /s
Energy: The energy photon contains can be calculate by this formula-- E=hf
f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.
So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20= 4.051*10^-19 J