Chapter 9, Problem 9.120QA

Interpretation Introduction

To find: a)How specific heat and  molar heat capacity data for the metals support the law of Dulong and Petit.

b) Use these data to predict the specific heat values of nickel and platinum. 

Answer to Problem 9.120QA

Solution:a)The product of the atomic mass of a metal times its specific heat equals to its molar heat capacity, therefore it is constant..

b) $specific heat of Nickel=0.420 J{/(g.}^{0}C)$

$specific heat of Platinum=0.127 J{/(g.}^{0}C)$

Explanation of Solution

Explanation: 1)    Concept:

Law of Dulong and Petit states that “the product of the atomic mass of a metal times its specific heat is approximately constant”.

Specific heat capacity is given by using formula,$c_{p,n}=\frac{q}{n\times ∆T}$

2) Given:

Metal in solid phase	${\mathit{c}}_{\mathit{p}}\mathit{ }\mathit{i}\mathit{n}\mathit{ }\mathit{J}{/(\mathit{g}.}^0\mathit{C})$	${\mathit{c}}_{\mathit{p},\mathit{n}}\mathit{ }\mathit{i}\mathit{n}\mathit{ }\mathit{J}{/(\mathit{m}\mathit{o}\mathit{l}.}^0\mathit{C})$
Aluminum	$0.897$	$24.2$
Carbon (graphite)	$0.71$	$8.5$
Carbon(diamond)	$0.127$	$1.5$
Chromium	$0.449$	$23.3$
Copper	$0.385$	$24.5$
Gold	$0.129$	$25.4$
Iron	$0.45$	$25.1$
Lead	$0.129$	$26.7$
Silicon	$0.71$	$19.9$
silver	$0.233$	$25.1$
Tin	$0.227$	$26.9$
Titanium	$0.523$	$25.0$
zinc	$0.387$	$25.3$

3) Calculation:

a)To calculate molar heat capacity of the metals, we use the specific heat of metal and atomic mass

For Aluminium:

$specific heat\times atomic mass=0.897J{/(g.}^{0}C)\times 26.98\frac{g}{mol}=24.2 J{/(mol.}^{0}C)$

For Chromium:

$specific heat\times atomic mass=0.449J{/(g.}^{0}C)\times 51.99\frac{g}{mol}=23.3 J{/(mol.}^{0}C)$

For Copper:

$specific heat\times atomic mass=0.385J{/(g.}^{0}C)\times 63.55\frac{g}{mol}=24.5 J{/(mol.}^{0}C)$

For gold:

$specific heat\times atomic mass=0.129J{/(g.}^{0}C)\times 196.97\frac{g}{mol}=25.4 J{/(mol.}^{0}C)$

For Iron:

$specific heat\times atomic mass=0.45J{/(g.}^{0}C)\times 55.85\frac{g}{mol}=25.1 J{/(mol.}^{0}C)$

For lead:

$specific heat\times atomic mass=0.129J{/(g.}^{0}C)\times 207.2\frac{g}{mol}=26.7 J{/(mol.}^{0}C)$

For Silicon:

$specific heat\times atomic mass=0.71J{/(g.}^{0}C)\times 28.09\frac{g}{mol}=19.9 J{/(mol.}^{0}C)$

For Silver:

$specific heat\times atomic mass=0.233J{/(g.}^{0}C)\times 107.87\frac{g}{mol}=25.1 J{/(mol.}^{0}C)$

For Tin:

$specific heat\times atomic mass=0.227J{/(g.}^{0}C)\times 118.71\frac{g}{mol}=26.9 J{/(mol.}^{0}C)$

For Titanium:

$specific heat\times atomic mass=0.523J{/(g.}^{0}C)\times 47.87\frac{g}{mol}=25.0 J{/(mol.}^{0}C)$

For Zinc:

$specific heat\times atomic mass=0.387J{/(g.}^{0}C)\times 65.38\frac{g}{mol}=25.3 J{/(mol.}^{0}C)$

The values of product of atomic mass of a metal and its specific heat are equal to its molar heat capacity. Molar heat capacity value for a metal is constant. Therefore, the Dulong and Petit law is true.

b) Take the average of all the above product values $=24.7J{/(mol.}^{0}C)$

From the above law, we can write that, $specific heat\times atomic mass=constant$

Therefore,$specific heat=\frac{constant}{atomic mass }$

Therefore,$specific heat of Nickel=\frac{24.7J{/(mol.}^{0}C)}{atomic mass of Nickel}=\frac{24.7J{/(mol.}^{0}C)}{58.69\frac{g}{mol}}=0.420 J{/(g.}^{0}C)$

$specific heat of Platinum=\frac{24.7J{/(mol.}^{0}C)}{atomic mass of Platinum}=\frac{24.7J{/(mol.}^{0}C)}{195.1\frac{g}{mol}}=0.127 J{/(g.}^{0}C)$

Conclusion:The product of the atomic mass of a metal times its specific heat is constant and equal to its molar heat capacity. From the given data and Dulong and Petit law, values of specific heat of nickel is $0.420 J{/(g.}^{0}C)$ and platinum is $0.127 J{/(g.}^{0}C)$.