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Physics · real student question

Solve for x: 1000 = (35)(0.02)(x − 293.15) + (0.6)(5.67e−8)(0.02)(x⁴ − 293.15⁴).

Question

A surface of area 0.02 m20.02\ \text{m}^2 loses 1000 W1000\ \text{W} by convection (coefficient 35 W/m2K35\ \text{W}/\text{m}^2\text{K}) plus radiation (emissivity 0.60.6, Stefan-Boltzmann constant 5.67×10−85.67 \times 10^{-8}) to surroundings at 293.15 K293.15\ \text{K}. Solve

1000=(35)(0.02)(x−293.15)+(0.6)(5.67×10−8)(0.02)(x4−293.154)1000 = (35)(0.02)(x - 293.15) + (0.6)(5.67\times10^{-8})(0.02)\left(x^4 - 293.15^4\right)

for the surface temperature xx in kelvin.

Step-by-step solution

  1. Collapse the constants into two coefficients. Group the convection and radiation factors:

    35×0.02=0.7,0.6×5.67×10−8×0.02=6.804×10−1035 \times 0.02 = 0.7, \qquad 0.6 \times 5.67\times10^{-8} \times 0.02 = 6.804\times10^{-10}

    so the equation is

    1000=0.7 (x−293.15)+6.804×10−10(x4−293.154)1000 = 0.7\,(x - 293.15) + 6.804\times10^{-10}\left(x^4 - 293.15^4\right)

  2. Expand into a quartic in standard form. With 293.154=7.38515×109293.15^4 = 7.38515\times10^{9}, the two constant contributions are 0.7×293.15=205.2050.7 \times 293.15 = 205.205 and 6.804×10−10×7.38515×109=5.02496.804\times10^{-10} \times 7.38515\times10^{9} = 5.0249:

    6.804×10−10 x4+0.7x−1210.23=06.804\times10^{-10}\,x^4 + 0.7x - 1210.23 = 0

  3. Do not stop at the linear estimate — this is the trap. Ignoring the quartic term gives

    x≈1210.230.7≈1728.9x \approx \frac{1210.23}{0.7} \approx 1728.9

    but substituting that back shows why it fails: 6.804×10−10(1728.9)4≈6079 W6.804\times10^{-10}(1728.9)^4 \approx 6079\ \text{W}, six times the entire heat load. The radiation term is dominant at that temperature, not negligible, so 1728.9 K1728.9\ \text{K} is far too high.

  4. Solve the quartic numerically instead. The left-hand side is strictly increasing in xx for x>0x > 0 (both 0.7x0.7x and x4x^4 increase), so there is exactly one physical root and bisection converges reliably. Bracketing between 300300 and 50005000 and bisecting gives

    x≈947.04 Kx \approx 947.04\ \text{K}

  5. Check the root by splitting the two loss mechanisms. At x=947.04 Kx = 947.04\ \text{K}:

    convection=0.7(947.04−293.15)=457.7 W\text{convection} = 0.7(947.04 - 293.15) = 457.7\ \text{W}

    radiation=6.804×10−10(947.044−293.154)=542.3 W\text{radiation} = 6.804\times10^{-10}\left(947.04^4 - 293.15^4\right) = 542.3\ \text{W}

    and 457.7+542.3=1000.0 W457.7 + 542.3 = 1000.0\ \text{W}, matching the required heat loss. Radiation carries 54%54\% of the load — confirming it could never have been dropped.

  6. State the answer with its physical reading.

    x≈947 K (≈674 ∘C)x \approx 947\ \text{K} \ (\approx 674\ ^\circ\text{C})

    The general lesson: whenever a balance mixes a linear convection term with a T4T^4 radiation term, the quartic must be solved, because radiation grows so steeply that it overtakes convection well before a few hundred kelvin of overtemperature.

Answer

x≈947 Kx \approx 947\ \text{K}

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