# Step-by-Step Solution ## **1. Problem Restatement** A 30-cm-diameter aluminum (Al 6061-T6) ball is heated from \( 70^\circ C \) to \( 180^\circ C \). - Density, \( \rho = 270 \, \text{kg/m}^3 \) - Specific heat, \( c = .896 \, \text{kJ/kg}^\circ C \) - Initial temperature, \( T_i = 70^\circ C \) - Final temperature, \( T_f = 180^\circ C \) - Diameter, \( D = .30 \, \text{m} \) **Find:** The amount of energy (\( Q \)) that needs to be transferred to the aluminum ball. --- ## **2. Schematic** ``` Q --------> [Aluminum Ball] (D = .30 m) Ti = 70°C Tf = 180°C ``` --- ## **3. List of Given/Assumed Parameters** | Parameter | Symbol | Value | |------------------------|----------|-------------------------| | Initial Temperature | \( T_i \) | \( 70^\circ C \) | | Final Temperature | \( T_f \) | \( 180^\circ C \) | | Diameter | \( D \) | \( .30 \) m | | Density | \( \rho \) | \( 270 \) kg/m³ | | Specific Heat | \( c \) | \( .896 \) kJ/kg·°C | --- ## **4. Formula for Energy Required** The energy required to increase the temperature is: \[ Q = m c \Delta T \] Where: - \( m \) = mass of the ball - \( c \) = specific heat - \( \Delta T = T_f - T_i \) --- ### **Step 1: Find the Mass of the Ball** The ball is a sphere: \[ \text{Volume} = V = \frac{4}{3} \pi r^3 \] \[ r = \frac{D}{2} = \frac{.30}{2} = .15 \, \text{m} \] \[ V = \frac{4}{3} \pi (.15)^3 = \frac{4}{3} \pi (.003375) \approx .01414 \, \text{m}^3 \] \[ m = \rho V = 270 \times .01414 \approx 38.18 \, \text{kg} \] --- ### **Step 2: Calculate the Temperature Change** \[ \Delta T = T_f - T_i = 180 - 70 = 110^\circ C \] --- ### **Step 3: Plug Values into the Formula** \[ Q = m c \Delta T \] > Note: \( c \) is given in kJ/kg·°C, so the answer will be in **kJ**. \[ Q = 38.18 \times .896 \times 110 \] \[ Q = 38.18 \times 98.56 \] \[ Q \approx 3765.2 \, \text{kJ} \] --- ## **5. Final Answer** \[ \boxed{Q = 376 \, \text{kJ}} \] (Rounded to 2 significant digits.) --- ### **Summary Table** | Quantity | Value | |------------------|---------------| | Mass (\( m \)) | 38.18 kg | | \( \Delta T \) | 110 °C | | \( Q \) | 376 kJ | --- ### **Conclusion** **The amount of energy that needs to be transferred to the aluminum ball is \(\boxed{376 \, \text{kJ}}\).**