To solve the problem, we need to determine the surface area of the vessel from which heat energy is being transferred. We can use the formula for heat transfer through conduction, which is given by Fourier's law of heat conduction: \[ Q = \frac{k \cdot A \cdot \Delta T}{d} \] Where: - \( Q \) is the heat transferred (in Joules), - \( k \) is the thermal conductivity of the material (in W/m·K), - \( A \) is the surface area through which heat is being transferred (in m²), - \( \Delta T \) is the temperature difference (in K), - \( d \) is the thickness of the material (in m). However, in this case, we are not given the thickness \( d \) of the vessel, but we can assume that the heat is being transferred through the surface area of the vessel directly into the surrounding environment. ### Step-by-Step Solution 1. **Identify the Given Values:** - Heat energy given off, \( Q = 1.2 \times 10^6 \, \text{J} \) - Time, \( t = 1 \, \text{s} \) - Thermal conductivity, \( k = 400 \, \text{W/m·K} \) - Temperature gradient, \( \Delta T = 30 \, \text{K/m} \) 2. **Calculate the Power:** Since the heat is given off in 1 second, we can calculate the power \( P \) (in Watts) using the formula: \[ P = \frac{Q}{t} = \frac{1.2 \times 10^6 \, \text{J}}{1 \, \text{s}} = 1.2 \times 10^6 \, \text{W} \] 3. **Rearranging Fourier's Law:** We can rearrange the formula to solve for the surface area \( A \): \[ A = \frac{Q \cdot d}{k \cdot \Delta T} \] However, since we are not given \( d \), we can express \( A \) in terms of power: \[ P = \frac{k \cdot A \cdot \Delta T}{d} \] Rearranging gives: \[ A = \frac{P \cdot d}{k \cdot \Delta T} \] 4. **Assuming a Unit Thickness:** For simplicity, we can assume \( d = 1 \, \text{m} \) (this is a common assumption when thickness is not specified). Thus, we can simplify our equation: \[ A = \frac{1.2 \times 10^6 \, \text{W} \cdot 1 \, \text{m}}{400 \, \text{W/m·K} \cdot 30 \, \text{K/m}} \] 5. **Calculating the Surface Area:** Now we can plug in the values: \[ A = \frac{1.2 \times 10^6}{400 \cdot 30} \] \[ A = \frac{1.2 \times 10^6}{12000} \] \[ A = 100 \, \text{m}^2 \] 6. **Final Calculation:** \[ A = 1.0 \times 10^2 \, \text{m}^2 \] ### Conclusion The correct answer is **A. 1.0 x 10² m²**. ### Explanation of Other Options: - **B. 9.0 x 10² m²**: This value is too high based on the calculations. It suggests a much larger surface area than what is required to dissipate the given amount of heat at the specified thermal conductivity and temperature gradient. - **C. 1.0 x 10³ m²**: This is also too high and does not align with the calculated area. It would imply an unrealistic scenario for the given parameters. - **D. 9.0 x 10⁴ m²**: This is excessively large and not feasible for the heat transfer described in the problem. ### Revision Summary: - Use Fourier's law of heat conduction to relate heat transfer, thermal conductivity, area, and temperature difference. - Calculate power from heat energy and time. - Rearrange the heat transfer equation to solve for surface area. - Always check units and assumptions (like thickness) when parameters are not provided.