Answer:
Dimension of specific heat will be [tex]=L^2T^{-2}\Theta ^{-1}[/tex]
Explanation:
We know that heat [tex]Q=mc\Delta T[/tex], Q is heat generated, m is mass, c is specific heat and [tex]\Delta T[/tex] is temperature difference
From formula we can write [tex]c=\frac{Q}{m\times \Delta T}[/tex]
Now unit of Q is joule or N-m
Newton can be written as [tex]kgm/sec^2[/tex]
So unit of Q is [tex]kgm^2/sec^2[/tex]
For dimension we use M for kg, L for meter(m) ,T for sec and [tex]\Theta[/tex] for temperature
So dimension of Q is [tex]ML^2T^{-2}[/tex]
So dimension of specific heat will be [tex]\frac{ML^2T^{-2}}{M\Theta }=L^2T^{-2}\Theta ^{-1}[/tex]
