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To find the surface area of the vessel, we use the formula for the rate of heat transfer by conduction:

Mathematics

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Answer

A. 1.0×102m2\text{A. } 1.0 \times 10^2 \text{m}^2

Step-by-step solution

4 steps

To find the surface area of the vessel, we use the formula for the rate of heat transfer by conduction:

Q=KAdTdxQ = K A \frac{dT}{dx}

where:

  • QQ is the rate of heat transfer (heat energy per unit time)
  • KK is the thermal conductivity
  • AA is the surface area
  • dTdx\frac{dT}{dx} is the temperature gradient

Given values:

  • Heat energy given off = 1.2×106 J1.2 \times 10^6 \text{ J}
  • Time = 1 s1 \text{ s}
  • Thermal Conductivity (KK) = 400 Wm−1K−1400 \text{ Wm}^{-1}\text{K}^{-1}
  • Temperature gradient (dTdx\frac{dT}{dx}) = 30 Km−130 \text{ Km}^{-1}
  1. Calculate the rate of heat transfer (QQ).

    The rate of heat transfer is the heat energy given off per second.

    Q=Heat energyTime=1.2×106 J1 s=1.2×106 WQ = \frac{\text{Heat energy}}{\text{Time}} = \frac{1.2 \times 10^6 \text{ J}}{1 \text{ s}} = 1.2 \times 10^6 \text{ W}
  2. Rearrange the formula to solve for the surface area (AA).

    A=QKdTdxA = \frac{Q}{K \frac{dT}{dx}}
  3. Substitute the given values into the rearranged formula.

    A=1.2×106 W400 Wm−1K−1×30 Km−1A = \frac{1.2 \times 10^6 \text{ W}}{400 \text{ Wm}^{-1}\text{K}^{-1} \times 30 \text{ Km}^{-1}}
  4. Calculate the surface area.

    A=1.2×10612000A = \frac{1.2 \times 10^6}{12000} A=1.2×1061.2×104A = \frac{1.2 \times 10^6}{1.2 \times 10^4} A=106−4A = 10^{6-4} A=102 m2A = 10^2 \text{ m}^2

    The surface area of the vessel is 1.0×102 m21.0 \times 10^2 \text{ m}^2.

    The final answer is A. 1.0×102m2\text{A. } 1.0 \times 10^2 \text{m}^2.

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