Respuesta :
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, [tex]T_{a} = 140^{\circ}C[/tex] = 273 + 140 = 413 K
Mass of Iron block, m = 36 kg
Temperature for iron block, [tex]T_{i} = 60^{\circ}C[/tex] = 273 + 60 = 333 K
At 400 k
Specific heat of Aluminium, [tex]C_{p} = 0.949\ kJ/kgK[/tex]
At room temperature
Specific heat of iron, [tex]C_{p} = 0.45\ kJ/kgK[/tex]
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
[tex]MC_{p}\Delta T = mC_{p}\Delta T[/tex]
[tex]28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)[/tex]
Thus
[tex]T_{e} = 109.71^{\circ}C[/tex] = 273 + 109.71 = 382.71 K
where
[tex]T_{e}[/tex] = Equilibrium temperature
Now,
To calculate the changer in entropy:
[tex]\Delta s = \Delta s_{a} + \Delta s_{i}[/tex]
Now,
For Aluminium:
[tex]\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}[/tex]
[tex]\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K[/tex]
For Iron:
[tex]\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}[/tex]
[tex]\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K[/tex]
Thus
[tex]\Delta s =-2.025 + 2.253 = 0.228\ kJ/K[/tex]
