Respuesta :
Answer:
[tex]k=105.0359\times 10^4\,W.m^{-1}.K^{-1}[/tex]
Explanation:
Given:
temperature at the hotter end, [tex]T_H=100^{\circ}C[/tex]
temperature at the cooler end, [tex]T_C=0^{\circ}C[/tex]
length of rod through which the heat travels, [tex]dx=0.7\,m[/tex]
cross-sectional area of rod, [tex]A=1.1\times 10^{-4}\,cm^2[/tex]
mass of ice melted at zero degree Celsius, [tex]m=8.7\times 10^{-3}\,kg[/tex]
time taken for the melting of ice, [tex]t=15\times60=900\,s[/tex]
thermal conductivity k=?
By Fourier's Law of conduction we have:
[tex]\dot{Q}=k.A.\frac{dT}{dx}[/tex]......................................(1)
where:
[tex]\dot{Q}[/tex]=rate of heat transfer
dT= temperature difference across the length dx
Now, we need the total heat transfer according to the condition:
we know the latent heat of fusion of ice, [tex]L = 334000\,J.kg^{-1}[/tex]
[tex]Q=m.L[/tex]
[tex]Q=8.7\times 10^{-3}\times 334000[/tex]
[tex]Q=2905.8\,J[/tex]
Now the heat rate:
[tex]\dot{Q}=\frac{Q}{t} [/tex]
[tex]\dot{Q}=\frac{2905.8}{900} [/tex]
[tex]\dot{Q}=3.2287\,W[/tex]
Now using eq,(1)
[tex]3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}[/tex]
[tex]k=205.4606\,W.m^{-1}.K^{-1}[/tex]
The thermal conductivity k of the metal is 68.93 W/mK.
Heat conducted by the ice
The quantity of heat conducted by the ice is calculated as follows;
Q = mcΔθ
Q = 8.7 x 4.184 x 100
Q = 3640 J
Heat heat flow rate
P = E/t = Q/t = 3640/(15 x 60) = 4.04 J/s = 4.04 W
Thermal conductivity
The thermal conductivity of the wire is calculated as follows;
P = kAΔT/x
Px = kAΔT
k = Px/AΔT
- ΔT = 100 - 0 = 100 C = 373 k
- A = 1.1 cm² = 1.1 x 10⁻⁴ m²
- x = 70 cm = 0.7 m
k = (4.04 x 0.7) / (373 x 1.1 x 10⁻⁴)
k = 68.93 W/mK
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