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Question

Question: The unit of thermal conductivity is: A. \(Jm{{K}^{-1}}\) B. \(J{{m}^{-1}}{{K}^{-1}}\) C. \(Wm{...

The unit of thermal conductivity is:
A. JmK1Jm{{K}^{-1}}
B. Jm1K1J{{m}^{-1}}{{K}^{-1}}
C. WmK1Wm{{K}^{-1}}
D. Wm1K1W{{m}^{-1}}{{K}^{-1}}

Explanation

Solution

Hint: First of all, the formula of thermal conductivity can be given as, q=kΔTLq=-k\dfrac{\Delta T}{L}. So, in this formula we will substitute the units of temperature, length and k and on the basis of that we will derive the unit of thermal conductivity.

Formula used: q=kΔTLq=-k\dfrac{\Delta T}{L}

Complete step-by-step answer:
In question we are asked to find the unit of thermal conductivity, so first of all, we will understand what thermal conductivity is.
So, thermal conductivity can be defined as heat flow rate across two walls at different temperatures separated at distance L from each other. This can be given mathematically as,
q=kΔTLq=-k\dfrac{\Delta T}{L}
Where, TT is temperature whose unit is KK, L is length whose unit is mm, k is thermal conductivity and q is heat flux whose unit is in W/m2W/{{m}^{2}}.
So, on substituting all these values in equation we will get,
Wm2=kKm\dfrac{W}{{{m}^{2}}}=-k\dfrac{K}{m}
Wmm2K=k\Rightarrow \dfrac{Wm}{{{m}^{2}}K}=k
Here, we will ignore negative sign as we are considering only units of the quantities so, on simplifying further we will get,
k=WmK\Rightarrow k=\dfrac{W}{mK}
It can also be written as k=Wm1K1k=W{{m}^{-1}}{{K}^{-1}}.
Hence, the unit of thermal conductivity is Wm1K1W{{m}^{-1}}{{K}^{-1}}.
Thus, option (d) is the correct answer.

Note: Here, the students might make mistake in considering the unit of q as W/m2W/{{m}^{2}} and instead of that they might consider J/m2J/{{m}^{2}}, but it is wrong and due to that the answer will also be wrong. So, students should always consider the units carefully and then solve the problems.