Using the equation for thermal conduction

PHYSICS · Multi-part · AP · Physics 2: Algebra-based · Thermodynamics · Thermodynamics

Problem

Using the equation for thermal conduction, calculate the following: (a) The rate of thermal energy transfer through a conductor of cross-sectional area 8 cm28\,\text{cm}^2 and thermal conductivity 200 J s−1 ∘C−1 m−1200\,\text{J s}^{-1}\,{}^\circ\text{C}^{-1}\text{ m}^{-1} if the temperature gradient across the conductor is 2.0 ∘C m−12.0\,{}^\circ\text{C m}^{-1}. (b) The cross-sectional area of a conductor of length 65 cm65\,\text{cm} and thermal conductivity 420 J s−1 ∘C−1 m−1420\,\text{J s}^{-1}\,{}^\circ\text{C}^{-1}\text{ m}^{-1} if a temperature difference of 40 ∘C40\,{}^\circ\text{C} between its ends makes thermal conduction occur at a rate of 52 W52\,\text{W}. (c) The temperature gradient necessary to cause conduction to occur at a rate of 3.5 W3.5\,\text{W} through a conductor of cross-sectional area 25 cm225\,\text{cm}^2 and thermal conductivity 40 J s−1 ∘C−1 m−140\,\text{J s}^{-1}\,{}^\circ\text{C}^{-1}\text{ m}^{-1}. (d) The thermal conductivity of glass if conduction occurs at a rate of 600 W600\,\text{W} through a window of length 2.0 m2.0\,\text{m}, height 1.6 m1.6\,\text{m}, and thickness 4.0 mm4.0\,\text{mm}, when the temperatures on either side of the window are 30 ∘C30\,{}^\circ\text{C} and 30.9 ∘C30.9\,{}^\circ\text{C}.

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