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The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.5 m. The block has a thermal conductivity of 200 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 37 ˚C and that of the cooler surface is 7 ˚C Determine the heat that flows in 4 s for each case.

Respuesta :

Answer:

[tex]Q_p=18000\ J[/tex]

[tex]Q_o=8000\ J[/tex]

[tex]Q_g=72000\ J[/tex]

Explanation:

Given:

  • Length of block, [tex]l=3L_o=3\times 0.5=1.5\ m[/tex]
  • Breadth of block, [tex]b=L_o=0.5\ m[/tex]
  • height of block, [tex]h=2L_o=2\times 0.5=1\ m[/tex]
  • Thermal conductivity of the block, [tex]k=200\ W.m.^{\circ}C[/tex]
  • Temperature on the hotter side, [tex]T_H=37^{\circ}C[/tex]
  • temperature on the cooler side, [tex]T_L=7^{\circ}C[/tex]
  • time for which the heat flows, [tex]t=4\ s[/tex]

REFER THE ATTACHED IMAGE FOR THE REFERENCE

The rate of heat flow using Fourier's law of conduction is given as:

[tex]\frac{Q}{t}=k.A.\frac{dT}{dx}[/tex]

Now the amount heat flow perpendicular to the pink surface:

[tex]\frac{Q_p}{4}=200\times (0.5\times 1.5).\frac{30}{1}[/tex]

[tex]Q_p=18000\ J[/tex]

Now the amount heat flow perpendicular to the orange surface:

[tex]\frac{Q_o}{4}=200\times (0.5\times 1).\frac{30}{1.5}[/tex]

[tex]Q_o=8000\ J[/tex]

Now the amount heat flow perpendicular to the green surface:

[tex]\frac{Q_g}{4}=200\times (1.5\times 1).\frac{30}{0.5}[/tex]

[tex]Q_g=72000\ J[/tex]

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