Sure! Let’s break down and solve the problem step by step: --- # **Problem Overview** We have **two spheres** of **10 kg** each: - **Sphere 1 (Brass):** - \( C_p = 1200 \) J/kg·K - \( k = 200 \) W/m·K - **Sphere 2 (Aluminum):** - Density \( \rho = 2707 \) kg/m³ - \( C_p = 800 \) J/kg·K - \( k = 200 \) W/m·K Both are cooled from **300°C (573 K)** to **100°C (373 K)** in water at **0°C (273 K)** The **convective heat transfer coefficient**: \( h = 80 \) W/m²·K The **characteristic length** for a sphere: \( L_c = \frac{r}{3} \) --- ## **Step 1: Find the characteristic length and surface area** ### **a. Volume and radius of the spheres** Since both spheres have the same mass, \( m = 10 \) kg: \[ \text{Volume} = \frac{m}{\rho} \] For **brass** (let’s assume \( \rho_{brass} \approx 8500 \) kg/m³): \[ V_{brass} = \frac{10}{8500} = 0.001176 \text{ m}^3 \] For **aluminum**: \[ V_{al} = \frac{10}{2707} = 0.003694 \text{ m}^3 \] \[ V = \frac{4}{3}\pi r^3 \implies r = \left( \frac{3V}{4\pi} \right)^{1/3} \] - **Brass**: \( r_{brass} = \left( \frac{3 \times 0.001176}{4\pi} \right)^{1/3} \approx 0.066 \) m - **Aluminum**: \( r_{al} = \left( \frac{3 \times 0.003694}{4\pi} \right)^{1/3} \approx 0.097 \) m #### **Surface area:** \[ A = 4\pi r^2 \] - \( A_{brass} \approx 0.0547 \) m² - \( A_{al} \approx 0.118 \) m² #### **Characteristic length:** \[ L_c = \frac{r}{3} \] --- ## **Step 2: Lumped Capacitance Method Check** The lumped capacitance method is valid if: \[ Bi = \frac{h L_c}{k} < 0.1 \] Let’s check for both spheres (\( k = 200 \) W/m·K): - **Brass:** \( Bi = \frac{80 \times 0.066/3}{200} = 0.0088 \) - **Aluminum:** \( Bi = \frac{80 \times 0.097/3}{200} = 0.013 \) Both \( Bi < 0.1 \), so **lumped capacitance applies**. --- ## **Step 3: Solving for cooling time** The lumped capacitance solution: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left( -\frac{hA}{\rho V C_p} t \right) \] We know: - \( T_i = 573 \) K - \( T_f = 373 \) K - \( T_\infty = 273 \) K So, \[ \frac{373 - 273}{573 - 273} = \exp\left( -\frac{hA}{\rho V C_p} t \right) \implies \frac{100}{300} = \exp\left( -\frac{hA}{\rho V C_p} t \right) \implies \frac{1}{3} = \exp(-b t) \] Take natural log: \[ \ln\left(\frac{1}{3}\right) = -b t \implies t = -\frac{\ln(1/3)}{b} \] Where \( b = \frac{hA}{m C_p} \) (since \( \rho V = m \)) --- ### **Calculate \( b \) for each sphere** #### **Brass:** \[ b_{brass} = \frac{80 \times 0.0547}{10 \times 1200} = \frac{4.376}{12000} = 0.0003647 \] #### **Aluminum:** \[ b_{al} = \frac{80 \times 0.118}{10 \times 800} = \frac{9.44}{8000} = 0.00118 \] --- ### **Calculate time for each:** \[ t = -\frac{\ln(1/3)}{b} = \frac{1.0986}{b} \] - **Brass:** \( t_{brass} = \frac{1.0986}{0.0003647} \approx 3013 \) s - **Aluminum:** \( t_{al} = \frac{1.0986}{0.00118} \approx 931 \) s --- ## **Conclusion** - **Which sphere cools faster?** **The aluminum sphere cools faster.** - **Why?** This is mainly because the aluminum sphere has a larger surface area and a lower heat capacity per unit mass, which allows it to lose heat more rapidly to the water. --- ### **Summary Table** | Sphere | Time to Cool (s) | |-------------|------------------| | Brass | ~3013 s | | Aluminum | ~931 s | --- ### **Associated Behavior** This behavior is associated with the **thermal properties** (heat capacity, density) and the **surface area** of the spheres. A higher surface area and lower heat capacity allow for faster cooling. --- If you have any questions about a particular step, let me know!