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Question:

A massless spring (k=800 N/m), attached with a mass (500g) is completely immersed in 1 kg of water. The spring is stretched by 2cm 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?

10⁻⁵K

10⁻³K

10⁻¹K

10⁻⁴K

Solution:

Correct option is D. 10⁻⁵K
By law of conservation of energy
Loss in spring potential energy = Gain in heat energy to mass and water.
1/2kx² = (m₁s₁ + m₂s₂)ΔT
ΔT = (1/2 * 800 N/m * (0.02 m)² ) / ( (0.5 kg * 420 J/kg·K) + (1 kg * 4200 J/kg·K) )
ΔT = 0.16 J / 4620 J/K
ΔT = 3.65 x 10⁻⁵ K