A massless spring (K=800 N/m), attached with a mass (500g) is completely immersed in 1 kg of water. The springs is stretched by 2cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibratioins stop completely ? (Assume that the water container and spring receive negligible heat and specific heat of mass =400 J/ kg K, specific heat of water =4184 J/kg K)

Options

A. `10^(-4) K`
B. `10^(-5) K`
C. `10^(-1) K`
D. `10^(-3) K`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy stored in the spring
The energy stored in a spring when it is stretched or compressed is given by the formula:
\[ E = \frac{1}{2} k x^2 \]
where:
- \( k = 800 \, \text{N/m} \) (spring constant)
- \( x = 2 \, \text{cm} = 0.02 \, \text{m} \) (displacement)

Substituting the values:
\[ E = \frac{1}{2} \times 800 \times (0.02)^2 \]
\[ E = \frac{1}{2} \times 800 \times 0.0004 = 0.16 \, \text{J} \]

### Step 2: Set up the energy conservation equation
The energy stored in the spring will be converted into heat energy that raises the temperature of both the mass and the water. The heat energy can be expressed as:
\[ Q = m_s c_s \Delta \theta + m_w c_w \Delta \theta \]
where:
- \( m_s = 0.5 \, \text{kg} \) (mass of the block)
- \( c_s = 400 \, \text{J/kg K} \) (specific heat of the block)
- \( m_w = 1 \, \text{kg} \) (mass of water)
- \( c_w = 4184 \, \text{J/kg K} \) (specific heat of water)
- \( \Delta \theta \) = change in temperature

### Step 3: Substitute the known values into the equation
Substituting the values into the energy equation:
\[ 0.16 = (0.5 \times 400 \Delta \theta) + (1 \times 4184 \Delta \theta) \]
\[ 0.16 = (200 \Delta \theta) + (4184 \Delta \theta) \]
\[ 0.16 = (200 + 4184) \Delta \theta \]
\[ 0.16 = 4384 \Delta \theta \]

### Step 4: Solve for \(\Delta \theta\)
Now, we can solve for \(\Delta \theta\):
\[ \Delta \theta = \frac{0.16}{4384} \]
Calculating this gives:
\[ \Delta \theta \approx 3.648 \times 10^{-5} \, \text{K} \]

### Step 5: Determine the order of magnitude
The order of magnitude of the change in temperature is:
\[ \Delta \theta \approx 3.7 \times 10^{-5} \, \text{K} \approx 10^{-5} \, \text{K} \]

### Final Answer
The order of magnitude of the change in temperature of the water when the vibrations stop completely is approximately \( 10^{-5} \, \text{K} \).

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