Let's break down the problem step by step: **Step 1: Data Listing** - **Iron**: Mass \( m_1 = 2 \) kg Initial Temp \( T_1 = 600 \) K Heat capacity \( c_1 = 0.460 \) kJ/(kg·K) - **Copper**: Mass \( m_2 = 3 \) kg Initial Temp \( T_2 = 800 \) K Heat capacity \( c_2 = 0.385 \) kJ/(kg·K) - **Granite**: Mass \( m_3 = 10 \) kg Initial Temp \( T_3 = 300 \) K Heat capacity \( c_3 = 0.790 \) kJ/(kg·K) **Step 2: Conservation of Energy** No heat exchange with the environment, so the **total internal energy of the system is conserved** (assuming negligible work is stored as kinetic, potential, etc. except for the "maximum work" we are to calculate). **Step 3: Concept** - Maximum work can be obtained if the process is **reversible** (i.e., the blocks are brought to a common final temperature \( T_f \) via a reversible process). - The **maximum work** is the difference between the decrease in the system's internal energy and the increase in its entropy multiplied by the final temperature. But since the final state must be equilibrium (all at \( T_f \)), let's use: \[ W_{max} = \Delta U_{initial} - \Delta U_{final} \] But since the system is isolated, **the actual energy lost from the blocks as they come to equilibrium is the maximum amount of work you could extract** (the rest goes to entropy increase). But the more precise method is to use the **exergy concept**, which for an isolated system, the maximum work is the decrease in the system's availability (exergy), and for a process where the final state is at a uniform temperature and no interaction with the environment, this is related to the entropy change. \[ W_{max} = U_{initial} - U_{final} - T_0 (S_{initial} - S_{final}) \] But since **no reservoir** is involved, and the system is isolated, the maximum work is the difference between the initial internal energy and the internal energy at final equilibrium. However, for the process to be reversible, the entropy of the system must not decrease. But in this case, the maximum work is the **decrease in the system's total exergy**, i.e., the difference between the actual change in internal energy and the minimum possible change in energy at no entropy production (reversible process). The most straightforward way is to say: \[ W_{max} = \left( \sum_i m_i c_i (T_{i, initial} - T_f) \right) \] But we must be careful: this is the total energy that could be lost **if all blocks go to \( T_f \)**, but the system must also satisfy entropy conservation in the reversible case. ### Step 4: Find Final Equilibrium Temperature (\(T_f\)) For a reversible process, the **final equilibrium temperature** is the one that **maximizes the entropy of the system**, but for a reversible process, entropy is conserved (the process is reversible, S stays constant). However, in reality, the maximum work is extracted if the final temperature is such that the **total entropy change of the system is zero** (since the process is reversible): \[ \Delta S_{total} = 0 \] For each block, the entropy change is: \[ \Delta S_i = m_i c_i \ln \left( \frac{T_f}{T_{i, initial}} \right) \] So, set the sum to zero: \[ \sum_{i=1}^3 m_i c_i \ln \left( \frac{T_f}{T_{i, initial}} \right) = 0 \] Plug in values: \[ 2 \times 0.460 \ln \left( \frac{T_f}{600} \right) + 3 \times 0.385 \ln \left( \frac{T_f}{800} \right) + 10 \times 0.790 \ln \left( \frac{T_f}{300} \right) = 0 \] Calculate products: - \( 2 \times 0.460 = 0.92 \) - \( 3 \times 0.385 = 1.155 \) - \( 10 \times 0.790 = 7.90 \) So, \[ 0.92 \ln \left( \frac{T_f}{600} \right) + 1.155 \ln \left( \frac{T_f}{800} \right) + 7.90 \ln \left( \frac{T_f}{300} \right) = 0 \] Combine logs: \[ \ln \left[ \left( \frac{T_f}{600} \right)^{0.92} \left( \frac{T_f}{800} \right)^{1.155} \left( \frac{T_f}{300} \right)^{7.90} \right] = 0 \] Exponentiate both sides: \[ \left( \frac{T_f}{600} \right)^{0.92} \left( \frac{T_f}{800} \right)^{1.155} \left( \frac{T_f}{300} \right)^{7.90} = 1 \] \[ T_f^{0.92 + 1.155 + 7.90} \cdot (600^{-0.92}) (800^{-1.155}) (300^{-7.90}) = 1 \] Sum exponents: \( 0.92 + 1.155 + 7.90 = 9.975 \) So, \[ T_f^{9.975} = 600^{0.92} \times 800^{1.155} \times 300^{7.90} \] \[ T_f = \left[ 600^{0.92} \times 800^{1.155} \times 300^{7.90} \right]^{1/9.975} \] Let's compute each term: First, calculate the logs to evaluate exponents: - \( \ln 600 = 6.39693 \) - \( \ln 800 = 6.68461 \) - \( \ln 300 = 5.70378 \) Now: - \( 600^{0.92} = e^{0.92 \cdot 6.39693} = e^{5.88518} = 361.4 \) - \( 800^{1.155} = e^{1.155 \cdot 6.68461} = e^{7.71982} = 2252.4 \) - \( 300^{7.90} = e^{7.90 \cdot 5.70378} = e^{45.060} = 2.67 \times 10^{19} \) Now multiply: \[ 361.4 \times 2252.4 = 814,522 \] \[ 814,522 \times 2.67 \times 10^{19} = 2.175 \times 10^{25} \] Now take the \( 1 / 9.975 \) root: \[ T_f = (2.175 \times 10^{25})^{1/9.975} \] Take log base 10 first: \[ \log_{10}(2.175 \times 10^{25}) = \log_{10} 2.175 + 25 = 0.337 + 25 = 25.337 \] So, \[ \log_{10}(T_f) = \frac{25.337}{9.975} = 2.541 \] So \[ T_f = 10^{2.541} = 347.5 \text{ K} \] Let's check with natural logs for more accuracy: \[ \ln(2.175 \times 10^{25}) = \ln(2.175) + 25 \ln 10 = 0.777 + 57.5646 = 58.3416 \] So \[ \ln(T_f) = \frac{58.3416}{9.975} = 5.849 \] So \[ T_f = e^{5.849} = 347.6 \text{ K} \] So, final equilibrium temperature: \[ \boxed{T_f = 348 \text{ K} \,\,\, (\text{rounded to 3 significant figures})} \] --- **Step 5: Calculate Maximum Work Extractable** The maximum work is the decrease in total internal energy from the initial state to the final equilibrium state: \[ W_{max} = \sum_{i=1}^3 m_i c_i (T_{i, initial} - T_f) \] Plug in the numbers: - **Iron:** \( 2 \times 0.460 \times (600 - 348) = 0.92 \times 252 = 231.84 \) kJ - **Copper:** \( 3 \times 0.385 \times (800 - 348) = 1.155 \times 452 = 521.46 \) kJ - **Granite:** \( 10 \times 0.790 \times (300 - 348) = 7.90 \times (-48) = -379.2 \) kJ Note: The granite block is absorbing heat (its temperature rises), so this is a negative contribution to available work. Sum up: \[ W_{max} = 231.84 + 521.46 - 379.2 = 374.1 \text{ kJ} \] --- **Step 6: Final State Description** - All blocks are at the same final temperature: \( T_f = 348 \) K. - The maximum work that can be extracted is **374 kJ** (rounded to 3 significant figures). --- ### **Final Answer** **Step-by-step summary:** 1. **Set up entropy balance:** \( 0.92 \ln \left( \frac{T_f}{600} \right) + 1.155 \ln \left( \frac{T_f}{800} \right) + 7.90 \ln \left( \frac{T_f}{300} \right) = 0 \) 2. **Solve for \( T_f \):** \( T_f = 348 \) K (rounded to 3 significant digits) 3. **Calculate maximum work:** \( W_{max} = 231.8 + 521.5 - 379.2 = 374 \) kJ (rounded) 4. **Final state:** All blocks at 348 K. \[ \boxed{ \begin{array}{l} \text{The maximum amount of work that can be extracted is } \mathbf{374\ \text{ kJ}}. \\ \text{In the final state, all blocks (iron, copper, granite) are at } \mathbf{348\ \text{K}}. \end{array} } \] --- **Summary Table:** | Block | Mass (kg) | \( c \) (kJ/kg·K) | \( T_{initial} \) (K) | \( T_{final} \) (K) | \( Q = m c (T_{i} - T_f) \) (kJ) | |---------|-----------|-------------------|-----------------------|---------------------|-----------------------------------| | Iron | 2 | 0.460 | 600 | 348 | 231.8 | | Copper | 3 | 0.385 | 800 | 348 | 521.5 | | Granite | 10 | 0.790 | 300 | 348 | -379.2 | | **Sum** | | | | | **374** | --- **Boxed Final Answer:** \[ \boxed{ \begin{array}{l} \text{The maximum work that can be extracted is } 374 \text{ kJ}. \\ \text{The final state is all three blocks at } 348 \text{ K}. \end{array} } \]