Answer:
note:
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Since there is no diagram I can see to prove my answer, I must infer that a heat source should be applied to the solid substance and cause the atoms in the solid to spread further apart and become a different state, in which atoms will be able to freely move, gas.
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
Mass is how heavy is it, weight is the size both are the same
Answer:
Explanation:
Given
mass of sled =26 kg
coefficient of static friction 
coefficient of kinetic friction 
In order to move sled from rest we need to provide a force greater than static friction which is given by

After Moving Sled kinetic friction comes in to play which is less than static friction

therefore minimum force to keep moving sledge at constant velocity is 18.34 N