Answer:
Dimension of specific heat will be
![=L^2T^(-2)\Theta ^(-1)](https://img.qammunity.org/2020/formulas/engineering/college/ju9qx85atj3jq7mugffh1yr5kbbgrj7aqb.png)
Step-by-step explanation:
We know that heat
, Q is heat generated, m is mass, c is specific heat and
is temperature difference
From formula we can write
![c=(Q)/(m* \Delta T)](https://img.qammunity.org/2020/formulas/engineering/college/gq49qe1x1hc4itcd6sgmutdlyubwgjhlir.png)
Now unit of Q is joule or N-m
Newton can be written as
![kgm/sec^2](https://img.qammunity.org/2020/formulas/engineering/college/1x91ckpcl5cuasq2gnugw15npkziqazhm7.png)
So unit of Q is
![kgm^2/sec^2](https://img.qammunity.org/2020/formulas/engineering/college/x4oeuyux1byfovo28s9acpzj6zvlugohau.png)
For dimension we use M for kg, L for meter(m) ,T for sec and
for temperature
So dimension of Q is
![ML^2T^(-2)](https://img.qammunity.org/2020/formulas/engineering/college/gbuw6h52g8b0rcd27jb1tijdbpvkgkdm3c.png)
So dimension of specific heat will be
![(ML^2T^(-2))/(M\Theta )=L^2T^(-2)\Theta ^(-1)](https://img.qammunity.org/2020/formulas/engineering/college/fudfnr2k0thh3chmpzx9xggvng23dq2ww4.png)