Let's go through each step in detail, showing the solution and providing explanations. --- # **Step 1: Specific Heat of Copper at 300 K** **Question:** What is the specific heat of copper at 300 K, in J/kg·K? **Solution:** This is a data lookup step. The specific heat of copper at room temperature (approximately 300 K) is well-known: \[ c_{\text{Cu}} \approx 385 \, \text{J/kg·K} \] **Final Answer:** \[ \boxed{385 \ \text{J/kg·K}} \] --- # **Step 2: Thermal Conductivity of Glass at 300 K** **Question:** What is the thermal conductivity of glass at 300 K, in W/m·K? **Solution:** Again, this is a data lookup step. The thermal conductivity of glass at around 300 K (room temperature) is: \[ k_{\text{glass}} \approx .78 \, \text{W/m·K} \] **Final Answer:** \[ \boxed{.78 \ \text{W/m·K}} \] --- # **Step 3: Rate of Heat Transfer Per Unit Area (Copper, 10 s)** **Question:** What is the rate of heat transfer per unit area 10 seconds after the finger touches copper, in W/m²? **Explanation and Formula:** For transient heat conduction between a semi-infinite solid (the plate) and another body (the finger), the heat flux is: \[ q = \frac{T_{\text{finger}} - T_{\text{plate}}}{\sqrt{\frac{\pi t}{\alpha \rho c}} \left(\frac{1}{\sqrt{k_{\text{finger}} \rho_{\text{finger}} c_{\text{finger}}}} + \frac{1}{\sqrt{k_{\text{plate}} \rho_{\text{plate}} c_{\text{plate}}}}\right)} \] But a standard formula for the initial heat flux into a semi-infinite solid is: \[ q = \frac{T_{\text{finger}} - T_{\text{plate}}}{\sqrt{\pi \alpha t}} \cdot \sqrt{k \rho c} \] Where: - \(T_{\text{finger}} = 310\, \text{K}\) - \(T_{\text{plate}} = 278\, \text{K}\) - \(t = 10\, \text{s}\) - \(k_{\text{Cu}} = 401\, \text{W/m·K}\) - \(\rho_{\text{Cu}} = 8933\, \text{kg/m}^3\) - \(c_{\text{Cu}} = 385\, \text{J/kg·K}\) First, calculate the thermal diffusivity: \[ \alpha_{\text{Cu}} = \frac{k_{\text{Cu}}}{\rho_{\text{Cu}} c_{\text{Cu}}} = \frac{401}{8933 \times 385} = \frac{401}{3,439,205} \approx 1.17 \times 10^{-4}\ \text{m}^2/\text{s} \] Now, calculate the heat flux: \[ (T_{\text{finger}} - T_{\text{plate}}) = 310 - 278 = 32\, \text{K} \] \[ \sqrt{\pi \alpha t} = \sqrt{\pi \times 1.17 \times 10^{-4} \times 10} = \sqrt{3.676 \times 10^{-3}} \approx .0606 \] \[ q = \frac{32}{.0606} \times \sqrt{401 \times 8933 \times 385} \] But this form is confusing; a more standard approach is: \[ q = \frac{(T_{\text{finger}} - T_{\text{plate}}) \, k}{\sqrt{\pi \alpha t}} \] So, \[ q = \frac{32 \times 401}{.0606} = \frac{12,832}{.0606} \approx 211,785\, \text{W/m}^2 \] **Final Answer:** \[ \boxed{2.1 \times 10^{5}\ \text{W/m}^2} \] --- # **Step 4: Contact Temperature (Finger + Copper)** **Formula:** The contact temperature between two semi-infinite solids is: \[ T_c = \frac{T_{1} \sqrt{k_1 \rho_1 c_1} + T_{2} \sqrt{k_2 \rho_2 c_2}} {\sqrt{k_1 \rho_1 c_1} + \sqrt{k_2 \rho_2 c_2}} \] Where (1) is the finger, (2) is copper. - For finger: \(T_1 = 310\,\text{K}\), \(\rho_1 = 100\), \(c_1 = 418\), \(k_1 = .625\) - For copper: \(T_2 = 278\,\text{K}\), \(\rho_2 = 8933\), \(c_2 = 385\), \(k_2 = 401\) \[ \sqrt{k_1 \rho_1 c_1} = \sqrt{.625 \times 100 \times 418} = \sqrt{2,612,500} \approx 1616 \] \[ \sqrt{k_2 \rho_2 c_2} = \sqrt{401 \times 8933 \times 385} = \sqrt{1,378,850,885} \approx 37,140 \] Plug into formula: \[ T_c = \frac{310 \times 1616 + 278 \times 37,140}{1616 + 37,140} = \frac{501,000 + 10,323,000}{38,756} = \frac{10,824,000}{38,756} \approx 279.3\,\text{K} \] **Final Answer:** \[ \boxed{279.3\ \text{K}} \] --- # **Step 5: Contact Temperature (Finger + Glass)** **Given:** - For glass: \(k_2 = .78\), \(\rho_2 = 223\), \(c_2 = 840\) - \(T_2 = 278\,\text{K}\) \[ \sqrt{k_2 \rho_2 c_2} = \sqrt{.78 \times 223 \times 840} = \sqrt{1,462,104} \approx 1209 \] \[ T_c = \frac{310 \times 1616 + 278 \times 1209}{1616 + 1209} = \frac{501,000 + 336,102}{2825} = \frac{837,102}{2825} \approx 296.5\,\text{K} \] **Final Answer:** \[ \boxed{296.5\ \text{K}} \] --- ## **Summary Table** | Step | Quantity | Value | |------|--------------------------------------------|-----------------------------| | 1 | Specific heat of copper (\(c_{\text{Cu}}\))| \(385\, \text{J/kg·K}\) | | 2 | Thermal conductivity of glass (\(k_{\text{glass}}\)) | \(.78\, \text{W/m·K}\) | | 3 | Heat transfer rate, copper (10 s) | \(2.1 \times 10^{5}\, \text{W/m}^2\) | | 4 | Contact temp (finger + copper) | \(279.3\, \text{K}\) | | 5 | Contact temp (finger + glass) | \(296.5\, \text{K}\) | --- ## **Explanation** - **Why does copper feel colder?** The contact temperature with copper is much lower than with glass, and the heat transfer rate is much higher, drawing heat away from your finger faster. --- Let me know if you need any calculation details or further clarification!