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sweet-ann [11.9K]

3 years ago

9

A cabin has a 0.159-m thick wooden floor [k = 0.141 W/(m · C°)] with an area of 13.4 m2. A roaring fire keeps the interior of th

e cabin at a comfortable 18.0 °C while the air temperature in the crawl space below the cabin is –16.4°C. What is the rate of heat conduction through the wooden floor?

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

[/tex] 408.8[/tex]

Explanation:

$t$ = thickness of the floor = 0.159 m

$k$ = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹

$T_{i}$ = Temperature of the interior of the cabin = 18.0 ⁰C

$T_{o}$ = Temperature of the air below the cabin = - 16.4 ⁰C

Difference in temperature is given as

$\Delta T$ = Difference in temperature = $T_{i} - T_{o} = 18 - (- 16.4) = 34.4$ ⁰C

$A$ = Area of the floor = 13.4 m²

$Q$ = Rate of heat conduction

Rate of heat conduction is given as

$Q = \frac{kA \Delta T}{t}$

$Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8$ W

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$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

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Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

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