Answer
$$
k=42.86 \mathrm{N} / \mathrm{m} .
$$
$$\left|\frac{d S}{d x}\right|=2.65 \times 10^{-3} \mathrm{J} / \mathrm{K} \cdot \mathrm{m}$$
Work Step by Step
Using Hooke's law, we find the spring constant to be
$$
k=\frac{F_{s}}{x_{s}}=\frac{1.50 \mathrm{N}}{0.0350 \mathrm{m}}=42.86 \mathrm{N} / \mathrm{m} .
$$
and by use Eq. $20-7$ we can get
$$\left|\frac{d S}{d x}\right|=\frac{k|x|}{T}=\frac{(42.86 \mathrm{N} / \mathrm{m})(0.0170 \mathrm{m})}{275 \mathrm{K}}=2.65 \times 10^{-3} \mathrm{J} / \mathrm{K} \cdot \mathrm{m}$$
