physics

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The average rate at which energy is conducted outward through the ground surface in a certain region is 51.6 mW/m2, and the average thermal conductivity of the near-surface rocks is 3.31 W/m·K. Assuming a surface temperature of 11.7°C, find the temperature (in Celsius) at a depth of 35.0 km (near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

  • physics -

    recall that for an energy transfer,
    q = -kA * dT / dx
    where
    q = rate of heat tranfer (in Watts)
    k = thermal conductivity (in W/m-K)
    dT = differential change in Temperature (in Celsius or in Kelvin)
    dx = differential change in position (x-direction)
    *note: this is negative since (T2-T1) is always negative, for heat always flows from higher temp to lower temp.

    assuming heat transfer is at steady state (does not change with time), this formula simplifies to
    q = -kA (T2-T1)/(x2-x1) ; or
    q/A = -k*(T2 - T2)/(x2 - x1)
    where
    T1 = heat source temp (at this problem it's the crust)
    T2 = lower temp (at this problem, it's the surface)
    first we convert 51.6 mW/m^2 into W by multiplying by 10^-3
    substituting,
    51.6 * 10^(-3) = -3.31*( 11.7 - T1 )/(35)
    0.0516*35/(-3.31) = 11.7 - T1
    -0.546 = 11.7 - T1
    -12.246 = -T1
    T1 = 12.246 deg C

    hope this helps~ :)

  • physics -

    oops, scrap my first answer,, i forgot to convert 35 km to meters:
    35 km = 35000 m

    substituting,
    51.6 * 10^(-3) = -3.31*( 11.7 - T1 )/(35000)
    0.0516*35000/(-3.31) = 11.7 - T1
    -545.62 = 11.7 - T1
    -557.32 = -T1
    T1 = 557.32 deg C

    sorry~ ^^;
    hope this helps~ :)

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