Question

A massless spring $(K=800{\text{Nm}}^{-1}),$ attached with a mass $\left(500\text{g}\right)$ is completely immersed in $1\text{kg}$ of water. The spring is stretched by $2\text{cm}$ and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely?
(Assume that the water container and spring receive negligible heat and specific heat of mass $=400{\text{J kg}}^{-1}{\text{K}}^{-1},$ specific heat of water $=4184{\text{J kg}}^{-1}{\text{K}}^{-1})$

• A. ${10}^{-4}\text{K}$
• B. ${10}^{-1}\text{K}$
• C. ${10}^{-5}\text{K}$
• D. ${10}^{-3}\text{K}$

Answer 1

The correct option is C ${10}^{-5}\text{K}$

Using law of conservation of energy,

$\frac{1}{2}k{x}^2=mC\Delta T+m_{\omega }{C}_{\omega }\Delta T$

$\frac{1}{2}\times 800\times (0.02{)}^2=0.5\times 400\times \Delta T+1\times 4184\times \Delta T$

$\therefore \Delta T=1\times {10}^{-5}\text{K}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.