Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
<span>Shading.
When light hits an opaque surface some is absorbed, the rest is reflected, The reflected light is called shading. Reflection is not simple and varies with material.
The surface’s structure defines the details of reflection. Variations produce anything from bright specular reflection</span>
Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
Explanation:
From the question we are told that
The time constant 
The potential across the capacitor can be mathematically represented as
Where 
is the voltage of the capacitor when it is fully charged
So at
Generally energy stored in a capacitor is mathematically represented as
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at
The energy stored can be evaluated at as
Hence the fraction of the energy stored in an initially uncharged capacitor is
