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Question: The amount of heat energy \( Q \) , used to heat substances depends on its mass \( m \) , its specif...

The amount of heat energy QQ , used to heat substances depends on its mass mm , its specific heat capacity SS and the change in temperature ΔT\Delta T of the substance. Using dimensional method, find expression for SS ( Given that [S]=[L2T−2K−1]\left[ S \right] = \left[ {{L^2}{T^{ - 2}}{K^{ - 1}}} \right] )
(A) QmΔTQm\Delta T
(B) QmΔT\dfrac{Q}{{m\Delta T}}
(C) QmΔT\dfrac{{Qm}}{{\Delta T}}
(D) mQΔT\dfrac{m}{{Q\Delta T}}

Explanation

Solution

Hint : From the dimension of energy and mass find the expression for specific heat using dimensional analysis.The specific heat capacity is the proportionality constant and its value depends on the nature of the substance.

Complete Step By Step Answer:
We have given here that, the amount of heat energy is proportional to the mass of the substance, change in temperature of the substance and it is proportional to the specific heat of the substance.
So, let the heat energy of used is related to the mass of the substance as, Q∝mxQ \propto {m^x}
it is related to the change in temperature as, Q∝ΔTyQ \propto \Delta {T^y}
and it is related to specific heat of the substance as, Q∝SzQ \propto {S^z}
So, we can write from the theorem of complex proportionality,
Q∝mx(ΔT)ySzQ \propto {m^x}{(\Delta T)^y}{S^z}
Q=kmx(ΔT)ySzQ = k{m^x}{(\Delta T)^y}{S^z} where, kk is dimensionless constant.
Now, we know from dimensional analysis, the dimension of the right hand side of the equation and the left hand side of the equation will be the same.
Now, dimension of heat energy is equal to dimension of energy, [Q]=[E]=[ML2T−2][Q] = [E] = [M{L^2}{T^{ - 2}}] and it is given that, [S]=[L2T−2K−1]\left[ S \right] = \left[ {{L^2}{T^{ - 2}}{K^{ - 1}}} \right]
So, we can write from dimensional analysis, [ML2T−2]=[Mx][Ky][L2zT−2zK−1z][M{L^2}{T^{ - 2}}] = [{M^x}][{K^y}][{L^{2z}}{T^{ - 2z}}{K^{ - 1z}}]
Or, [ML2T−2]=[MxL2zT−2zKy−z][M{L^2}{T^{ - 2}}] = [{M^x}{L^{2z}}{T^{ - 2z}}{K^{y - z}}]
Hence, comparing the power of both sides we can write, x=1x = 1 ,
2z=22z = 2
Or, z=1z = 1
y−z=0y - z = 0
Or, y=zy = z
So, y=1y = 1
Hence, we get the heat absorbed as, Q=m1(ΔT)1S1Q = {m^1}{(\Delta T)^1}{S^1} taking the value of constant coefficient as k=1k = 1 for convenience.
Or, Q=m(ΔT)SQ = m(\Delta T)S
So, the expression of specific heat capacity becomes, S=Qm(ΔT)S = \dfrac{Q}{{m(\Delta T)}}
Hence, option (B ) is correct.

Note :
Here, we have taken the value of the dimensionless quantity one as per our convenience. The heat absorbed by a substance is directly proportional to the mass of the object and change in temperature.