a spring of force constant k=8.40 x 10^3 N/m is compressed by 0.220m. It is placed into a vessel containing 3.70kg of water and then released.Assuming all the energy from the spring goes into heating the water, find the change in temperature of the water. Specific heat capacity of water is 4186J/(kg.K)?

Find the elastic potential energy ([tex]\text{EPE}[/tex]) stored in the spring using the formula [tex](\text{EPE}) = (1/2)\, k\, x^{2}[/tex], where [tex]k[/tex] is the spring constant and [tex]x[/tex] is the displacement of the spring.
Find the temperature change based on the heat capacity of the water in the vessel: [tex]\Delta T = Q / (c\, m)[/tex], where [tex]Q[/tex] is the energy absorbed, [tex]c[/tex] is the specific heat capacity of water, and [tex]m[/tex] is the mass of water in the vessel.

The elastic potential energy stored in this spring is:

[tex]\begin{aligned} (\text{EPE}) &= \frac{1}{2}\, k\, x^{2} \\ &= \frac{1}{2}\, (8.40 \times 10^{3}\; {\rm N\cdot m^{-1}})\, (0.220\; {\rm m})^{2} \\ &\approx 203.28\; {\rm J}\end{aligned}[/tex].

Assume that water in the vessel absorbed all these energy. Given the mass and heat capacity of water in the vessel, the temperature change would be:

[tex]\begin{aligned} \Delta T &= \frac{Q}{c\, m} \\ &\approx \frac{(203.28\; {\rm J})}{(4186\; {\rm J\cdot kg^{-1} \cdot K^{-1}})\, (3.70\; {\rm kg})} \\ &\approx 0.013\; {\rm K}\end{aligned}[/tex].

In other words, the temperature of water in the vessel would increase by approximately [tex]0.013\; {\rm K}[/tex] ([tex]0.013[/tex] degrees celsius.)