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, ![C_{p} = 0.949\ kJ/kgK](https://tex.z-dn.net/?f=C_%7Bp%7D%20%3D%200.949%5C%20kJ%2FkgK)
At room temperature
Specific heat of iron, ![C_{p} = 0.45\ kJ/kgK](https://tex.z-dn.net/?f=C_%7Bp%7D%20%3D%200.45%5C%20kJ%2FkgK)
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
![MC_{p}\Delta T = mC_{p}\Delta T](https://tex.z-dn.net/?f=MC_%7Bp%7D%5CDelta%20T%20%3D%20mC_%7Bp%7D%5CDelta%20T)
![28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)](https://tex.z-dn.net/?f=28%5Ctimes%200.949%28140%20-%20T_%7Be%7D%29%20%3D%2036%5Ctimes%200.45%28T_%7Be%7D%20-%2060%29)
Thus
= 273 + 109.71 = 382.71 K
where
= Equilibrium temperature
Now,
To calculate the changer in entropy:
![\Delta s = \Delta s_{a} + \Delta s_{i}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20%5CDelta%20s_%7Ba%7D%20%2B%20%5CDelta%20s_%7Bi%7D)
Now,
For Aluminium:
![\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}](https://tex.z-dn.net/?f=%5CDelta%20s_%7Ba%7D%20%3D%20MC_%7Bp%7Dln%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bi%7D%7D)
![\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K](https://tex.z-dn.net/?f=%5CDelta%20s_%7Ba%7D%20%3D%2028%5Ctimes%200.949%5Ctimes%20ln%5Cfrac%7B382.71%7D%7B413%7D%20%3D%20-%202.025%5C%20kJ%2FK)
For Iron:
![\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}](https://tex.z-dn.net/?f=%5CDelta%20s_%7Bi%7D%20%3D%20mC_%7Bi%7Dln%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bi%7D%7D)
![\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K](https://tex.z-dn.net/?f=%5CDelta%20s_%7Ba%7D%20%3D%2036%5Ctimes%200.45%5Ctimes%20ln%5Cfrac%7B382.71%7D%7B333%7D%20%3D%202.253%5C%20kJ%2FK)
Thus
![\Delta s =-2.025 + 2.253 = 0.228\ kJ/K](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D-2.025%20%2B%202.253%20%3D%200.228%5C%20kJ%2FK)