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Question: Dimensions of specific heat are: \(\text{A}\text{.}\left[ M{{L}^{2}}{{T}^{-2}}K \right]\) \(\tex...

Dimensions of specific heat are:
A.[ML2T−2K]\text{A}\text{.}\left[ M{{L}^{2}}{{T}^{-2}}K \right]
B.[ML2T−2K−1]\text{B}\text{.}\left[ M{{L}^{2}}{{T}^{-2}}{{K}^{-1}} \right]
C.[ML2T2K−1]\text{C}\text{.}\left[ M{{L}^{2}}{{T}^{2}}{{K}^{-1}} \right]
D.[L2T−2K−1]\text{D}\text{.}\left[ {{L}^{2}}{{T}^{-2}}{{K}^{-1}} \right]

Explanation

Solution

Hint: Convert the derived physical quantities into fundamental physical quantities. Specific heat depends on the heat applied, mass of the object and the rise in temperature of the object. So we can find the dimensional formula of specific heat from the dimension of these three quantities.

Complete step by step answer:
All the derived physical quantities can be expressed in terms of the fundamental quantities. The derived units are dependent on the 7 fundamental quantities. Fundamental units are mutually independent of each other.
Dimension of a physical quantity is the power to which the fundamental quantities are raised to express that physical quantity.
Now, Specific heat can be defined as the amount of heat needed to be supplied to material of unit mass to raise the temperature by one degree.
Specific heat(c)=Heat(H) × [Mass(M) × Temperature(K)](−1)\text{Specific heat(c)}=\text{Heat(H) }\times \text{ }{{[\text{Mass(M) }\times \text{ Temperature(K)}]}^{(-1)}}
Mass and Temperature are fundamental quantities.
The dimensional formula of Mass and Temperature are[M1L0T0]\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]and [M0L0T0K1]\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}}{{K}^{1}} \right]
Now, Heat is a derived quantity. We have to express in terms of the fundamental quantities.
Now, the dimension of heat energy is the same as the dimension of work done.
Work=Force×displacement\text{Work}=\text{Force}\times \text{displacement}
Force=mass×acceleration\text{Force}=\text{mass}\times \text{acceleration}
Acceleration=displacement×time(−2)\text{Acceleration}=\text{displacement}\times \text{tim}{{\text{e}}^{(-2)}}
Now. dimension of acceleration= [M0L1T−2]\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]
Dimension of work=[M1]×[M0L1T−2]×[L1]=[M1L2T−2][{{M}^{1}}]\times [{{M}^{0}}{{L}^{1}}{{T}^{-2}}]\times [{{L}^{1}}]=[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]
Hence, dimension of heat = [M1L2T−2]\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]
Dimension of specific heat = [M1L2T−2]×[M1]×[K1]−1=[M(1−1)L2T−2K−1]=[L2T−2K−1][{{M}^{1}}{{L}^{2}}{{T}^{-2}}]\times [{{M}^{1}}]\times {{[{{K}^{1}}]}^{-1}}=[{{M}^{(1-1)}}{{L}^{2}}{{T}^{-2}}{{K}^{-1}}]=[{{L}^{2}}{{T}^{-2}}{{K}^{-1}}]
So, the correct answer is option (D).

Note: Don’t try to remember the dimensional formula. You may get confused. Always express the derived quantities in terms of the fundamental quantities and you will get the dimension of quantities.