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
(C) 10⁻¹ K (D) 10⁻⁵ K

Answer 1

By law of conservation of energy

21 kx2=(m1s1+m2s2)ΔT

ΔT=438416×10−2=3.65×10−5.

Answer 2

The order of magnitude of the change in temperature is $10^{-5}$ K .

Option (D) is correct

Explanation:

Given :

Spring constant $k = 800 \frac{N}{m}$

Mass of object $m_{1} = 0.5$ Kg

Mass of water $m _{2} =$ 1 Kg

Displacement $x = 2 \times 10^{-2}$ m

Specific heat of object $c_{1} = 400 \frac{J}{kg K}$

Specific heat of water  $c_{2} = 4184 \frac{J}{kg K}$

According to the law of conservation,

Spring potential energy = Heat

    $\frac{1}{2} k x^{2} = (m_{1}c_{1} + m_{2} c_{2} ) \Delta T$

Where $\Delta T =$ change in temperature

     $\Delta T = \frac{1600 \times 10^{-4} }{ 4384} = 0.364 \times 10^{-4}$

Since order of magnitude is,

     $\Delta T = 3.64 \times 10^{-5}$

Thus, the order of magnitude of the change in temperature is $10^{-5}$ K