Rituraj Tiwari
Last Activity: 4 Years ago
To solve this problem, we will apply the principles of thermal equilibrium and entropy change calculations.
### Given data:
- Mass of aluminum, \( m_{Al} = 200 \) g = \( 0.2 \) kg
- Specific heat capacity of aluminum, \( c_{Al} = 900 \) J/kg·K
- Initial temperature of aluminum, \( T_{Al, i} = 100^\circ C = 373 \) K
- Mass of water, \( m_w = 50 \) g = \( 0.05 \) kg
- Specific heat capacity of water, \( c_w = 4186 \) J/kg·K
- Initial temperature of water, \( T_{w, i} = 20^\circ C = 293 \) K
The system is thermally isolated, meaning no heat is lost to the surroundings. The heat lost by aluminum will be gained by water until they reach thermal equilibrium.
### (a) Finding the equilibrium temperature
Using the principle of conservation of energy:
Heat lost by aluminum = Heat gained by water
\[
m_{Al} c_{Al} (T_{Al, i} - T_f) = m_w c_w (T_f - T_{w, i})
\]
Substituting the values:
\[
(0.2)(900)(373 - T_f) = (0.05)(4186)(T_f - 293)
\]
\[
180 (373 - T_f) = 209.3 (T_f - 293)
\]
Expanding:
\[
67140 - 180 T_f = 209.3 T_f - 61224.9
\]
\[
67140 + 61224.9 = 180 T_f + 209.3 T_f
\]
\[
128364.9 = 389.3 T_f
\]
\[
T_f = \frac{128364.9}{389.3} = 329.7 K
\]
Thus, the equilibrium temperature is **329.7 K** (approximately **57°C**).
---
### (b) Entropy change of aluminum
The entropy change is given by:
\[
\Delta S = m c \ln \left(\frac{T_f}{T_i}\right)
\]
For aluminum:
\[
\Delta S_{Al} = (0.2)(900) \ln \left(\frac{329.7}{373}\right)
\]
\[
\Delta S_{Al} = 180 \ln (0.8842)
\]
\[
\Delta S_{Al} = 180 (-0.1228)
\]
\[
\Delta S_{Al} = -22.1 \text{ J/K}
\]
---
### (c) Entropy change of water
For water:
\[
\Delta S_w = (0.05)(4186) \ln \left(\frac{329.7}{293}\right)
\]
\[
\Delta S_w = 209.3 \ln (1.125)
\]
\[
\Delta S_w = 209.3 (0.1178)
\]
\[
\Delta S_w = 24.7 \text{ J/K}
\]
---
### (d) Total entropy change of the system
\[
\Delta S_{\text{total}} = \Delta S_{Al} + \Delta S_w
\]
\[
\Delta S_{\text{total}} = -22.1 + 24.7
\]
\[
\Delta S_{\text{total}} = 2.6 \text{ J/K}
\]
Since the total entropy change is **positive**, this process is spontaneous and follows the second law of thermodynamics.
---
### Final Answers:
(a) The equilibrium temperature is **329.7 K (57°C)**.
(b) The entropy change of aluminum is **-22.1 J/K**.
(c) The entropy change of water is **24.7 J/K**.
(d) The total entropy change of the system is **2.6 J/K**.
