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 300 N/m is stretched by 10.0 cm, placed in 40 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.

Answer 1

This question is all about energy conservation:

Here,x = 10 cm = 0.10m

Potential Energy stored in spring = 1/2 x 300 N/m x (0.1m)2 = 1.5 J

now both of them will get heated and also they 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 = 40g x 1 cal/g oC x (T-18 oC)  

40g x 1 cal/g oC x (T-18 oC) + 20g x 0.30 cal/g oC x (T-18 oC) = 1.5

1 J = 0.239 cal

1.5 J = 0.3585 cal

(T-18 oC) (40 + 6) = 0.179253 cal

T-18 oC = 3.8968*10^-3 oC

T = 18.004 oC