Question

The molar heat capacity of rock salt at low temperatures varies with temperature according to Dobey's $T^3$ law. Thus $C=k\frac{T^3}{{\theta }^3}k=1940Jmo{l}^{-1}{K}^{-1}\theta =281k$.
How much heat is required to raise the temperature of $2$ moles of rock salt from $10k$ to $50K$ ?

• A. $800J$
• B. $373J$
• C. $273J$
• D. None of these

Answer 1

The correct option is C $273J$

Heat added due to increase in temperature by $dt$ is given by :

$dH=NCdT$

where $N$ is no. of moles present

Integrating the above equation :

$H=2\int CdT$

$H=2\int k\frac{T^3}{{281}^3}dT$

$H=2\times \frac{1940}{{281}^3}\times {\left[\frac{T^4}{4}\right]}_{10}^{50}$

$H=273$