A block of mass `500g` and specific heat `400J//KgK` is attached with a spring of spring constant `800 N//m`. Now block is dipped in water of mass `1kg` and specific heat `4184J//KgK`. Now the spring is elongated by `2cm` and released. Find rise in temperature of water and block system when block finally comes to rest.

Options

A. `7.64xx10^(-4)K`
B. `3.64xx10^(-3)K`
C. `3.64xx10^(-5)K`
D. `3.64xx10^(-6)K`

Answer

To solve the problem, we need to calculate the rise in temperature of the water and block system when the block comes to rest after being released from the elongated spring. Here are the steps to find the solution:

### Step 1: Understand the Energy Conversion
When the block is released, the potential energy stored in the spring will be converted into heat energy, which will raise the temperature of both the block and the water.

### Step 2: Calculate the Potential Energy of the Spring
The potential energy (PE) stored in the spring when it is elongated can be calculated using the formula:
\[ PE = \frac{1}{2} k a^2 \]
where:
- \( k = 800 \, \text{N/m} \) (spring constant)
- \( a = 0.02 \, \text{m} \) (elongation)

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

### Step 3: Set Up the Heat Energy Equation
The heat energy gained by the block and water when the block comes to rest is given by:
\[ Q = m_1 s_1 \Delta T + m_2 s_2 \Delta T \]
where:
- \( m_1 = 0.5 \, \text{kg} \) (mass of the block)
- \( s_1 = 400 \, \text{J/(kg K)} \) (specific heat of the block)
- \( m_2 = 1 \, \text{kg} \) (mass of the water)
- \( s_2 = 4184 \, \text{J/(kg K)} \) (specific heat of the water)
- \( \Delta T \) is the rise in temperature (which we need to find)

### Step 4: Substitute Values into the Heat Energy Equation
Now we can substitute the values into the equation:
\[ 0.16 = (0.5 \times 400 \Delta T) + (1 \times 4184 \Delta T) \]
This simplifies to:
\[ 0.16 = (200 \Delta T) + (4184 \Delta T) \]
\[ 0.16 = (200 + 4184) \Delta T \]
\[ 0.16 = 4384 \Delta T \]

### Step 5: Solve for \(\Delta T\)
Now, we can solve for \(\Delta T\):
\[ \Delta T = \frac{0.16}{4384} \]
Calculating this gives:
\[ \Delta T \approx 3.64 \times 10^{-5} \, \text{K} \]

### Final Answer
The rise in temperature of the water and block system when the block finally comes to rest is approximately:
\[ \Delta T \approx 3.64 \times 10^{-5} \, \text{K} \]