Question

The potential energy stored in a spring is
1/2*kx ^{2}*, where

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

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