Question

using euipartion of energy, the specific heat `("in" jkg^(-1)K^(-1)`of aluminium at room temperature can be estimated to be (atomic weigh of aluminium=27)
A. 410
B. 25
C. 1850
D. 925

Answer 1

Correct Answer - D
using law of equipartition of energy ,the heat developed in aluminium is
`2xx(3)/(2)K_BT=msDeltaT=ms(T-0)`
`rArr" " 3K_(B)=ms=(M)/(N_(A))s`
`rArr" " s=(3K_(B)N_(A))/(M)` (i)
`"Here", " " M=27,K_(B)=1.38xx10^(-23)JK^(-1)`
`"and " N_(A)=6.02xx10^(23)//"mole" `
from Eq (i), specific heat of aluminium at room temperature,
`s=(3xx1.38xx10^(-23)xx6.02xx10^(23))/(27xx10^(-3))" " [because1g=10^(-3)kg]`
`s=(138xx60.2)/(9)=923.06~~JKg^(-1)K^(-1)`