To solve the problem, we need to follow these steps:
### Step 1: Identify the given data
- Spring constant, \( k = 800 \, \text{N/m} \)
- Mass of the block, \( m = 500 \, \text{g} = 0.5 \, \text{kg} \)
- Mass of water, \( m_w = 1 \, \text{kg} \)
- Specific heat of the block (mass), \( s_m = 400 \, \text{J/kg K} \)
- Specific heat of water, \( s_w = 4184 \, \text{J/kg K} \)
- Extension of the spring, \( x = 2 \, \text{cm} = 0.02 \, \text{m} \)
### Step 2: Calculate the potential energy stored in the spring
The potential energy stored in the spring when it is stretched is given by the formula:
\[
PE = \frac{1}{2} k x^2
\]
Substituting the values:
\[
PE = \frac{1}{2} \times 800 \, \text{N/m} \times (0.02 \, \text{m})^2
\]
\[
PE = \frac{1}{2} \times 800 \times 0.0004 = 0.16 \, \text{J}
\]
### Step 3: Relate the potential energy to the heat energy absorbed by the water
When the spring stops vibrating, the potential energy is converted into heat energy absorbed by the water and the block. The total heat energy absorbed can be expressed as:
\[
Q = m_w s_w \Delta T + m s_m \Delta T
\]
Where \( \Delta T \) is the change in temperature of the water and the block.
### Step 4: Substitute the known values into the heat energy equation
\[
0.16 = (1 \, \text{kg} \times 4184 \, \text{J/kg K} + 0.5 \, \text{kg} \times 400 \, \text{J/kg K}) \Delta T
\]
Calculating the terms inside the parentheses:
\[
= (4184 + 200) \Delta T = 4384 \Delta T
\]
Thus, we have:
\[
0.16 = 4384 \Delta T
\]
### Step 5: Solve for \( \Delta T \)
\[
\Delta T = \frac{0.16}{4384}
\]
Calculating \( \Delta T \):
\[
\Delta T \approx 0.0000365 \, \text{K} \approx 3.65 \times 10^{-5} \, \text{K}
\]
### Step 6: Determine the order of magnitude of the change in temperature
The order of magnitude of \( \Delta T \) is \( 10^{-5} \, \text{K} \).
### Final Answer
The order of magnitude of the change in temperature of the water when the vibrations stop completely is \( 10^{-5} \, \text{K} \).
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