A massless spring (k = 800 N/m), attached with a mass (500 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 J/kg K, specific heat of water = 4184 J/kg K)

Question

A massless spring (k = 800 N/m), attached with a mass (500 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 J/kg K, specific heat of water = 4184 J/kg K) (A) 10-3 K  (B) 10-4 K  (C) 10-1 K  (D) 10-5 K

Tardigrade Admin · Accepted Answer

By law of conservation of energy (1/2) kx2 = (m1s1+ m2 s2) Δ T Δ T = (16 × 10-2/4384) = 3.65 × 10-5 .

$1 0^{- 3} K$

$1 0^{- 4} K$

$1 0^{- 1} K$

$1 0^{- 5} K$

By law of conservation of energy
$\frac{1}{2} k x^{2} = (m_{1} s_{1} + m_{2} s_{2}) Δ T$
$Δ T = \frac{16 \times 1 0 ^{- 2}}{4384} = 3.65 \times 1 0^{- 5}$ .

10−3K

10−4K

10−1K

10−5K

By law of conservation of energy 21​kx2=(m1​s1​+m2​s2​)ΔT ΔT=438416×10−2​=3.65×10−5 .

$1 0^{- 3} K$

$1 0^{- 4} K$

$1 0^{- 1} K$

$1 0^{- 5} K$

By law of conservation of energy
$\frac{1}{2} k x^{2} = (m_{1} s_{1} + m_{2} s_{2}) Δ T$
$Δ T = \frac{16 \times 1 0 ^{- 2}}{4384} = 3.65 \times 1 0^{- 5}$ .