A massless spring ($k = 800\ \text{N/m}$), attached with a mass ($500\ \text{g}$) 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 $= 400\ \text{J/kg K}$, specific heat of water $= 4184\ \text{J/kg K}$)
Answer: (A) $10^{-5}\ \text{K}$
All the stored spring energy ends up as heat in the water and the mass:
$$E = \frac12kx^2 = \frac12\times800\times(0.02)^2 = 0.16\ \text{J}$$
$$\Delta T = \frac{E}{m_wc_w + m c} = \frac{0.16}{1\times4184 + 0.5\times400} = \frac{0.16}{4384} \approx 3.6\times10^{-5}\ \text{K}$$
This is of the order of $10^{-5}\ \text{K}$.
Solution by Sreeraj P, M.Sc Physics