1. Initial energy of oscillation

When the spring-mass system is stretched by amplitude $A = 2\text{ cm} = 0.02\text{ m}$ and released, the total mechanical energy is

$E = \frac{1}{2}kA^2$

Given:

• $k = 800\,\text{N/m}$
• $A = 0.02\,\text{m}$

So,

$E = \frac{1}{2}(800)(0.02)^2$

$E = 400 \times 0.0004 = 0.16\,\text{J}$

2. Where does this energy go?

When the vibrations stop completely, this mechanical energy is dissipated as heat in the surrounding water.

The question says the spring and container receive negligible heat, so we take all this heat to go into the water.

Thus,

$Q \approx 0.16\,\text{J}$

3. Temperature rise of water

For water,

• mass $m_w = 1\,\text{kg}$
• specific heat $c_w = 4184\,\text{J/kg K}$

Using

$Q = m c \Delta T$

we get

$\Delta T = \frac{Q}{m_wc_w} = \frac{0.16}{1 \times 4184}$

$\Delta T \approx 3.8 \times 10^{-5}\,\text{K}$

4. Order of magnitude

$\Delta T \sim 10^{-5}\,\text{K}$

So the correct option is:

C: $10^{-5}\,\text{K}$

5. About the specific heat of the mass

The given specific heat of the mass is not needed here, because the problem asks for the change in temperature of water and explicitly states that the spring/container receive negligible heat. The oscillation energy is tiny and is taken to heat the water.

Therefore, the answer remains unchanged.