Question

The potential energy stored in a spring is 1/2kx², where k is the force constant of the spring and x is the distance the spring is stretched from equilibrium. Suppose a spring with force constant 150 N/m is stretched by 10.0 cm, placed in 10 g of water in an adiabatic container, and released. The mass of the spring is 20 g, and its specific heat capacity is 0.30 cal/(g °C). The initial temperature of the water and the spring is 18.000°C. The water’s specific heat capacity is 1.00 cal/(g °C). Find the final temperature of the water in degrees Celsius. Write down 5 significant figures

Confused with the equations I need to use, any help would be Appreciated!

Answer 1

You just need to do energy conservation:

x = 10 cm = 0.10m

P.E stored in spring = 1/2 x 150 N/m x (0.1m)² = 0.75 J

now both water and spring will get heated and will be at same temperature

energy absorbed by spring = mCpΔT = 20g x 0.30 cal/g ^oC x (T-18 ^oC)

energy absorbed by water = 10g x 1 cal/g ^oC x (T-18 ^oC)  

10g x 1 cal/g ^oC x (T-18 ^oC) + 20g x 0.30 cal/g ^oC x (T-18 ^oC) = 0.75 J

1 J = 0.239 cal

0.75 J = 0.179253 cal

(T-18 ^oC) (10 + 6) = 0.179253 cal

T-18 ^oC = 0.046875 ^oC

T = 18.0468 ^oC till 5 significant figures = 18.047 ^oC