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Answer:
Equilibrium Temperature is 382.71 K
Total entropy is 0.228 kJ/K
Solution:
As per the question:
Mass of the Aluminium block, M = 28 kg
Initial temperature of aluminium, = 273 + 140 = 413 K
Mass of Iron block, m = 36 kg
Temperature for iron block, = 273 + 60 = 333 K
At 400 k
Specific heat of Aluminium,
At room temperature
Specific heat of iron,
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
Thus
= 273 + 109.71 = 382.71 K
where
= Equilibrium temperature
Now,
To calculate the changer in entropy:
Now,
For Aluminium:
For Iron:
Thus
Answer:
W = 112 lb
Explanation:
Given:
- δb = 0.025 in
- E = 29000 ksi (A-36)
- Area A_de = 0.002 in^2
Find:
Compute Weight W attached at C
Solution:
- Use proportion to determine δd:
δd/5 = δb/3
δd = (5/3) * 0.025
δd = 0.0417 in
- Compute εde i.e strain in DE:
εde = δd / Lde
εde = 0.0417 / 3*12
εde = 0.00116
- Compute stress in DE, σde:
σde = E*εde
σde = 29000*0.00116
σde = 33.56 ksi
- Compute the Force F_de:
F_de = σde *A_de
F_de = 33.56*0.002
F_de = 0.0672 kips
- Equilibrium conditions apply:
(M)_a = 0
3*W - 5*F_de = 0
W = (5/3)*F_de
W = (5/3)* 0.0672 = 112 lb
