Let's break down the problem and solve step by step. --- ## **Step 1: Restate Key Information** - **Surface temperature, \( T_s = 57.0^\circ C \)** - **Body (initial) temperature, \( T_b = 37.0^\circ C \)** - **Critical temperature for cell death, \( T_c = 47.0^\circ C \)** - **Maximum allowed time above \( T_c \): \( \Delta t = 10 \) s** - **Thickness of epidermis, \( x = 1.0 \) mm \( = 1.0 \times 10^{-3} \) m** - **Thermal diffusivity, \( \alpha = 1.5 \times 10^{-7} \) m²/s** - **Specific heat, \( C_p = 4.0 \times 10^3 \) J/(kg·K)** --- ## **Step 2: Which Equation to Use?** This is a **transient heat conduction** problem with a sudden change in surface temperature. We want to know **how long it takes for the temperature at \( x = 1.0 \) mm to reach \( 47^\circ C \)**. For a semi-infinite solid with initial temperature \( T_b \), suddenly exposed to \( T_s \) at the surface at \( t = 0 \): \[ \frac{T(x,t) - T_s}{T_b - T_s} = \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \] Or rewritten: \[ T(x, t) = T_s + (T_b - T_s) \cdot \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \] We want to find \( t \) such that \( T(x, t) = T_c \). --- ## **Step 3: Plug in the Numbers** Rearrange for the error function: \[ T_c = T_s + (T_b - T_s) \cdot \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \] \[ \frac{T_c - T_s}{T_b - T_s} = \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \] Plug in values: - \( T_c = 47 \) - \( T_s = 57 \) - \( T_b = 37 \) \[ \frac{47 - 57}{37 - 57} = \frac{-10}{-20} = 0.5 \] So, \[ \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) = 0.5 \] The value for which \( \text{erf}(z) = 0.5 \) is \( z \approx 0.4769 \). --- ## **Step 4: Solve for \( t \)** \[ \frac{x}{2\sqrt{\alpha t}} = 0.4769 \] \[ x = 2 \cdot 0.4769 \cdot \sqrt{\alpha t} \] \[ \frac{x}{0.9538} = \sqrt{\alpha t} \] \[ \left(\frac{x}{0.9538}\right)^2 = \alpha t \] \[ t = \frac{1}{\alpha} \left(\frac{x}{0.9538}\right)^2 \] Plug in numbers: - \( x = 1.0 \times 10^{-3} \) m - \( \alpha = 1.5 \times 10^{-7} \) m²/s \[ t = \frac{1}{1.5 \times 10^{-7}} \left(\frac{1.0 \times 10^{-3}}{0.9538}\right)^2 \] Calculate inside the parenthesis: \[ \frac{1.0 \times 10^{-3}}{0.9538} = 1.048 \times 10^{-3} \] \[ (1.048 \times 10^{-3})^2 = 1.098 \times 10^{-6} \] Now for \( t \): \[ t = \frac{1.098 \times 10^{-6}}{1.5 \times 10^{-7}} \] \[ t = 7.319 \, \text{s} \] --- ## **Step 5: Interpret the Result** **It takes about 7.3 seconds for the temperature at a depth of 1 mm to reach 47°C.** But cell damage only occurs if the temperature is above 47°C for more than 10 s. Since it takes 7.3 s to reach 47°C, the **maximum allowed time for the instrument to be on is \( 7.3 + 10 = 17.3 \) seconds**. --- ## **## Final Answer** ### **To avoid unacceptable damage to the patient's skin, your instrument should collect diagnostic information within approximately \(\boxed{17 \, \text{seconds}}\).** --- ### **Summary of Steps** 1. **Used transient heat conduction (semi-infinite solid) solution.** 2. **Solved for time to reach critical temperature at 1 mm depth.** 3. **Added the extra allowed 10 s above threshold.** 4. **Final safe exposure time is about 17 seconds.** If you need more details or explanations for any step, let me know!