Question

If 277 J is required to change the temperature of 7.11 g of sodium chloride from 275.0 K to 335.0 K, what is the specific heat capacity of sodium chloride?

A. -0.650 J/g×K
B. 1.54 J/g×K
C. 1.18×105 J/g×K
D. 0.650 J/g×K
E. 6.50×10-4 J/g×K

Answer 1

$q=cm\Delta T$

$\therefore c= \frac{q}{m\Delta T}$

$c=\frac{277 J}{7.11 g \times (335.0 K-275.0 K)} = 0.650 \frac{J}{g \times K}$

Answer: D. $0.650 \frac{J}{g \times K}$