- #1

vcsharp2003

- 897

- 177

## Homework Statement

A spring (k = 500 N/m) supports a 400 g mass which is immersed in 900 g of water. The specific heat of the mass is 450 J/kg and of water is 4184 J/kg. The spring is now stretched 15 cm and, after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped?

I am ending up with ΔT ( change in temperature of thermodynamic system) and ΔU

_{int}(change in internal energy of thermodynamic system) as unknowns and one equation only, so it seems it's not possible to solve for temperature change ΔT.

## Homework Equations

heat = mass x specific heat X temperature change

## The Attempt at a Solution

[/B]

Let's consider the spring + water as a closed system. Also, let's look at the system starting from when the spring is in water and mass hanging from it in an equilibrium position to when mass returns to it's original position and vibration of spring has stopped. Then, for this system the following points will be true.

- there could be some heat flow across the system's boundary $$\Delta Q_{in} \neq 0$$

- the system does not do any work on it's surrounds nor is work done on the system $$W_{by} = 0$$

- the internal energy of the system should increase as it's temperature will rise $$\Delta U_{int} > 0$$
- the change in KE of the system is zero since it has no translational motion $$\Delta KE = 0 $$
- the change in PE of the system is change in spring PE + change in gravitational PE ( change in gravitational PE is zero but change in spring PE is not zero since the spring is less stretched at end of interval being considered i.e. spring PE has decreased)

$$(0 - 0) + (0 - \frac{1}{2} \times k \times x^{2}) + \Delta U_{int} = Q_{in} - 0$$

$$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = Q_{in} - 0$$

Also, by using the second formula under

**Relevant Equations**using specific heat we get another equation. If ΔT = temperature change in K over the time interval being considered, then

$$Q_{in} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T$$

If we combine above two equations then we get the following equation.

$$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T - 0$$.

It's impossible to solve for ΔT since we have two unknowns in above equation. I am not sure if I am applying first law of Thermodynamics correctly.

*NOTE*: However, if Q

_{in}is assumed to be zero since heat flows are happening within the system due to fluid resistance against vibration, and ΔU

_{int}= 0.9 x 4184 x ΔT + 0.4 x 450 ΔT then by substituting into first law of thermodynamics equation, we get only one unknown of ΔT which we can solve. But, I am not sure if this second approach is correct.