Question:
A massless spring \((k=800\ \mathrm{N/m})\), attached with a mass \((500\ \mathrm{g})\) is completely immersed in \(1\) kg of water. The spring is stretched by \(2\) 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 \(=400\ \mathrm{J/kg\,K}\), specific heat of water \(=4184\ \mathrm{J/kg\,K}\))
A massless spring \((k=800\ \mathrm{N/m})\), attached with a mass \((500\ \mathrm{g})\) is completely immersed in \(1\) kg of water. The spring is stretched by \(2\) 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 \(=400\ \mathrm{J/kg\,K}\), specific heat of water \(=4184\ \mathrm{J/kg\,K}\))
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The stored spring energy \(\frac12kx^2\) is finally converted into heat shared by the water and the attached mass.
Updated On: Nov 5, 2026
- \(10^{-3}\ \mathrm{K}\)
- \(10^{-1}\ \mathrm{K}\)
- \(10^{-5}\ \mathrm{K}\)
- \(10^{-4}\ \mathrm{K}\)
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The Correct Option is C
Solution and Explanation
Energy initially stored in the spring: \(E=\frac{1}{2}kx^2=\frac{1}{2}\times800\times(0.02)^2=0.16\) J.
All of it is dissipated as heat when the vibrations stop. Heat capacity of water plus mass: \(1\times4184+0.5\times400=4384\) J/K.
\(\Delta T=\frac{0.16}{4384}\approx3.65\times10^{-5}\) K.
Order of magnitude: \(10^{-5}\) K.
Answer: (3) \(10^{-5}\ \mathrm{K}\)
All of it is dissipated as heat when the vibrations stop. Heat capacity of water plus mass: \(1\times4184+0.5\times400=4384\) J/K.
\(\Delta T=\frac{0.16}{4384}\approx3.65\times10^{-5}\) K.
Order of magnitude: \(10^{-5}\) K.
Answer: (3) \(10^{-5}\ \mathrm{K}\)
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