Question:
A metal rod of length $10 \,cm$ and area of cross-section $2.8 \times 10^{-4} m ^{2}$ is covered with a non-conducting substance. One end of it is maintained at $80^{\circ} C$, while the other end is put in ice at $0^{\circ} C .$ It is found that $20\, gm$ of ice melts in $5\, min$. The thermal conductivity of the metal in $Js ^{-1} m ^{-1} K ^{-1}$ is (Latent heat of ice is $80 \,cal\, g ^{-1} .$ )
A metal rod of length $10 \,cm$ and area of cross-section $2.8 \times 10^{-4} m ^{2}$ is covered with a non-conducting substance. One end of it is maintained at $80^{\circ} C$, while the other end is put in ice at $0^{\circ} C .$ It is found that $20\, gm$ of ice melts in $5\, min$. The thermal conductivity of the metal in $Js ^{-1} m ^{-1} K ^{-1}$ is (Latent heat of ice is $80 \,cal\, g ^{-1} .$ )
Updated On: Apr 4, 2024
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The Correct Option is D
Solution and Explanation
Given,
length of rod, $l=10 \,cm =0.1 \,m$
area of cross-section of rod, $A=2.8 \times 10^{-4}\, m ^{2}$
temperatuer at one end, $T_{1}=80^{\circ} C$
temperature at other end, $T_{2}=0^{\circ} C$
quantity of melted ice, $m =20 \,gm$
time taken to melt ice, $=5 \,min =300 \,sec$
and latent heat of ice, $s=80\, cal \,g ^{-1}$
Now, rate of the heat flow $=\frac{m \times s \times 4.184}{t}$
$=\frac{20 \times 80 \times 4.184}{300}=22.314 J / s$
Rate of heat flow in the rod, $\Delta Q=\frac{k A \cdot \Delta T}{l}$
$22.314=\frac{k\left(2.8 \times 10^{4}\right) \times 80}{0.1} $
$\therefore k=99.61 \approx 100\, Js ^{-1} \,m ^{-1} \,K ^{-1}$
length of rod, $l=10 \,cm =0.1 \,m$
area of cross-section of rod, $A=2.8 \times 10^{-4}\, m ^{2}$
temperatuer at one end, $T_{1}=80^{\circ} C$
temperature at other end, $T_{2}=0^{\circ} C$
quantity of melted ice, $m =20 \,gm$
time taken to melt ice, $=5 \,min =300 \,sec$
and latent heat of ice, $s=80\, cal \,g ^{-1}$
Now, rate of the heat flow $=\frac{m \times s \times 4.184}{t}$
$=\frac{20 \times 80 \times 4.184}{300}=22.314 J / s$
Rate of heat flow in the rod, $\Delta Q=\frac{k A \cdot \Delta T}{l}$
$22.314=\frac{k\left(2.8 \times 10^{4}\right) \times 80}{0.1} $
$\therefore k=99.61 \approx 100\, Js ^{-1} \,m ^{-1} \,K ^{-1}$
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Concepts Used:
Heat Transfer
What is Heat Transfer?
It is defined as the movement of heat across the border of the system due to a difference in temperature between system and its surroundings.
How is Heat Transferred?
Heat can travel from one place to another in several ways. The different modes of heat transfer include:
- Conduction - Heat flows from things with higher temp to objects with lower temp.
- Convection - Movement of liquid molecules from higher temp regions to lower temp regions.
- Radiation - Radiant heat is present in every other way in our daily lives. Thermal radiations are also known to as radiant heat. Thermal radiation is generated by the emission of electromagnetic waves.
