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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. 410B. 25C. 1850D. 925 |
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Answer» 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)` |
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