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A 30-kg aluminum block initially at 1408c is brought into contact with a 40-kg block of iron at 608c in an insulated enclosure. determine the final equilibrium temperature and the total entropy change for this process.

Respuesta :

Aluminum:
m₁ = 30 kg, mass
T₁ = 1408 C = 1408 +273 = 1681 K
c₁ = 921 J/(kg-K), from tables

Iron:
m₂ = 40 kg
T₂ = 608 C = 881 K
c₂ = 460.5 J/(kg-K), from tables

Let the final temperature be T.
Because there are no heat losses, the First Law requires that heat loss from the aluminum be equal to heat gain by the iron.
That is,
(30 kg)*(921 J/(kg-K))*(1681 - T K) = (40 kg)*(460.5 J/(kg-K))*(T - 881 K)
27630(1681 - T) = 18420(T - 881)
1.5(1681) - 1.5T = T - 881
3.4025 x 10³ = 2.5T
T = 1361 K = 1088 C

Heat loss by the aluminum is
Q₁ = 30*921*(1088-1408) = - 8.8416 x 10⁶ J
Heat gained b the iron is
Q₂ = 40*460.5*(1088-608) = + 8.8416 x 10⁶ J

The entropy change of the aluminum is
S₁ = Q₁/T₁ + Q₁/T = (-8.8416 x 10⁶) *(1/1681 + 1/1361) = - 1.1756 x 10⁴ J/K
The ntropy change of the iron is
S₂ = Q₂/T₂ + Q₂/T = (8.8416 x 10⁶)*(1/881 + 1/1361) = 1.6532 x 10⁴ J/K

The entropy change for the process is
S₁+S₂ = 4776.1 J/K
The entropy change is positive, in accordance with the Second Law of Thermodynamics.

Answer:
Final temperature = 1088 C
Entropy change = 4776 J/K

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