Let's tackle **Part c step by step**. **Restatement:** If 18 MJ of energy is used to heat 150 kg of sulphuric acid at constant pressure from 10°C, what will be the final temperature? Use Newton-Raphson. --- **Step 1: Write the Energy Balance** Given: - Energy supplied, \( Q = 18\,\text{MJ} = 18,000,000\,\text{J} \) - Mass, \( m = 150\,\text{kg} \) - Initial temperature, \( T_1 = 10^\circ\text{C} \) - Final temperature, \( T_2 = ? \) - Need to find \( T_2 \) The enthalpy change: \[ Q = m \int_{T_1}^{T_2} c_p\,dT \] --- **Step 2: Express \( c_p \) in Consistent Units** Given: \[ \hat{C}_P = 23.06 + 2.071 \times 10^{-2}T \quad (\text{Btu/lb-mol}^\circ\text{R}) \] Let's convert to \( \text{kJ/kg}^\circ\text{C} \): - 1 Btu = 1.05506 kJ - 1 lb-mol \( \ce{H2SO4} \) = 98.08 lb = 44.454 kg So, \[ \text{Molar mass} = 98.08\,\text{g/mol} = 0.09808\,\text{kg/mol} \] But 1 lb-mol = 453.59237 g = 0.45359237 kg But since c_p is per lb-mol, first convert to per kg. **Conversion:** \[ \frac{\text{Btu}}{\text{lb-mol} \cdot ^\circ\text{R}} \times \frac{1.05506\,\text{kJ}}{1\,\text{Btu}} \times \frac{1\,\text{lb-mol}}{44.454\,\text{kg}} \times \frac{1\,^\circ\text{R}}{1\,^\circ\text{C}} \] But \( 1\,^\circ\text{C} = 1.8\,^\circ\text{R} \implies 1\,^\circ\text{R} = 5/9\,^\circ\text{C} \) So, \[ 1\,^\circ\text{R} = \frac{5}{9}\,^\circ\text{C} \] Thus, \[ \frac{\text{Btu}}{\text{lb-mol}^\circ\text{R}} \times \frac{1.05506}{44.454} \times \frac{9}{5} \quad \text{(to get kJ/kg}^\circ\text{C}) \] Calculate this conversion factor: \[ \frac{1.05506}{44.454} \approx 0.02374 \] \[ 0.02374 \times \frac{9}{5} = 0.02374 \times 1.8 = 0.04273 \] So, \[ \hat{C}_P (\text{Btu/lb-mol}^\circ\text{R}) \longrightarrow c_p (\text{kJ/kg}^\circ\text{C}) = \hat{C}_P \times 0.04273 \] --- **Step 3: Write the Integral** Given \( \hat{C}_P = 23.06 + 2.071 \times 10^{-2} T \) (with \( T \) in \( ^\circ\text{R} \)), and \( T \) (in \( ^\circ\text{R} \)) = \( t(^\circ\text{C}) \times 1.8 + 491.67 \). Let \( t \) (temperature in \( ^\circ\text{C} \)), so \( T = 1.8t + 491.67 \). \[ \hat{C}_P = 23.06 + 2.071 \times 10^{-2}(1.8t + 491.67) \] \[ = 23.06 + 0.02071 \times 1.8 t + 0.02071 \times 491.67 \] \[ = 23.06 + 0.03728 t + 10.189 \] \[ = (23.06 + 10.189) + 0.03728 t \] \[ = 33.249 + 0.03728 t \] So, \[ c_p (\text{kJ/kg}^\circ\text{C}) = (33.249 + 0.03728 t) \times 0.04273 \] \[ = 1.421 + 0.001594 t \qquad \text{(rounded to 4 decimals)} \] --- **Step 4: Set Up the Equation** The enthalpy change is: \[ Q = m \int_{T_1}^{T_2} (1.421 + 0.001594 t) \, dt \] \[ = m \left[1.421 (t_2 - t_1) + 0.001594 \frac{(t_2^2 - t_1^2)}{2}\right] \] Plug in values (\( t_1 = 10 \), \( m = 150 \), \( Q = 18,000\,\text{kJ} \)): \[ 18,000 = 150 \left[1.421 (t_2 - 10) + 0.001594 \frac{t_2^2 - 100}{2}\right] \] \[ 18,000 = 150 \left[1.421 (t_2 - 10) + 0.000797 (t_2^2 - 100)\right] \] \[ \frac{18,000}{150} = 1.421 (t_2 - 10) + 0.000797 (t_2^2 - 100) \] \[ 120 = 1.421 t_2 - 14.21 + 0.000797 t_2^2 - 0.0797 \] \[ 120 + 14.21 + 0.0797 = 1.421 t_2 + 0.000797 t_2^2 \] \[ 134.2097 = 1.421 t_2 + 0.000797 t_2^2 \] --- **Step 5: Rearrange for Newton-Raphson** Let \( f(t_2) = 0.000797 t_2^2 + 1.421 t_2 - 134.2097 \). We need to solve this for \( t_2 \). --- **Step 6: Newton-Raphson Iteration** Newton-Raphson formula: \[ t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)} \] Where: \[ f(t_2) = 0.000797 t_2^2 + 1.421 t_2 - 134.2097 \] \[ f'(t_2) = 2 \times 0.000797 t_2 + 1.421 = 0.001594 t_2 + 1.421 \] Let's guess \( t_2 \). Since original is 10°C, and 75°C in part b used about 10.7 MJ, let's guess \( t_2 \) is about 90°C. **First iteration: \( t_2^{(0)} = 90 \)°C** Calculate \( f(90) \): \[ f(90) = 0.000797 \times 90^2 + 1.421 \times 90 - 134.2097 \] \[ = 0.000797 \times 8100 + 127.89 - 134.2097 \] \[ = 6.4557 + 127.89 - 134.2097 \] \[ = 134.3457 - 134.2097 = 0.136 \] Now, \( f'(90) = 0.001594 \times 90 + 1.421 = 0.14346 + 1.421 = 1.56446 \) Apply Newton-Raphson: \[ t_1 = 90 - \frac{0.136}{1.5645} = 90 - 0.0869 = 89.91 \] **Second iteration: \( t_2^{(1)} = 89.91 \)°C** \[ f(89.91) = 0.000797 \times (89.91)^2 + 1.421 \times 89.91 - 134.2097 \] \( (89.91)^2 = 8083.21 \) \[ = 0.000797 \times 8083.21 + 1.421 \times 89.91 - 134.2097 \] \[ = 6.444 + 127.794 - 134.2097 \] \[ = 134.238 - 134.2097 = 0.0283 \] \( f'(89.91) = 0.001594 \times 89.91 + 1.421 = 0.1433 + 1.421 = 1.5643 \) \[ t_2 = 89.91 - \frac{0.0283}{1.5643} = 89.91 - 0.0181 = 89.892 \] **Third iteration: \( t_2^{(2)} = 89.892 \)°C** \[ f(89.892) = 0.000797 \times (89.892^2) + 1.421 \times 89.892 - 134.2097 \] \( (89.892)^2 = 8082.52 \) \[ = 0.000797 \times 8082.52 + 1.421 \times 89.892 - 134.2097 \] \[ = 6.443 + 127.791 - 134.2097 = 134.234 - 134.2097 = 0.0243 \] \( f'(89.892) = 0.001594 \times 89.892 + 1.421 = 0.14329 + 1.421 = 1.5643 \) \[ t_3 = 89.892 - \frac{0.0243}{1.5643} = 89.892 - 0.0155 = 89.876 \] But since the change is now very small, this is basically converged. --- ### **Final Answer** **The final temperature of the sulphuric acid is:** \[ \boxed{89.9^\circ\text{C}} \] --- ### **Summary of Steps** 1. Convert the given heat capacity equation to appropriate units and in terms of \( t \) in \( ^\circ\text{C} \). 2. Set up the energy balance equation. 3. Rearranged to a form suitable for Newton-Raphson. 4. Used Newton-Raphson iteration to solve for the final temperature. 5. Final answer is **\( 89.9^\circ\text{C} \)**. If you have any questions about any step, let me know!