A thin smooth tube of length 2 m containing a small pallet of mercury in it is rapidly inverted 50 times. If capillary effect is neglected find the approximate increase in temperature of the mercury. Given that the specific heat of mercury is 30 cal/kg K.

Answer

To solve the problem, we need to calculate the increase in temperature of the mercury after it has been inverted 50 times in a thin tube of length 2 m. We will use the principles of potential energy and specific heat.

### Step-by-Step Solution:

1. **Identify the potential energy change**:
When the tube is inverted, the mercury moves up and down. The change in height (h) for each inversion is the length of the tube, which is 2 m. The potential energy change (PE) for one inversion can be calculated using the formula:
\[ PE = m \cdot g \cdot h \]
where:
- \( m \) = mass of mercury (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height change (2 m)

2. **Calculate total potential energy change after 50 inversions**:
Since the mercury is inverted 50 times, the total potential energy change (PE_total) will be:
\[ PE_{\text{total}} = 50 \cdot (m \cdot g \cdot h) = 50 \cdot (m \cdot 9.81 \cdot 2) \]
Simplifying this gives:
\[ PE_{\text{total}} = 50 \cdot m \cdot 19.62 \]

3. **Relate potential energy to temperature change**:
The change in thermal energy (Q) of the mercury can be expressed using the specific heat formula:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( s \) = specific heat of mercury (30 cal/kg K)
- \( \Delta T \) = change in temperature (in K)

4. **Set the potential energy equal to thermal energy**:
Since the energy gained by the mercury is equal to the potential energy lost, we can set the two equations equal:
\[ 50 \cdot m \cdot 19.62 = m \cdot 30 \cdot \Delta T \]

5. **Cancel mass (m)**:
Since mass appears on both sides of the equation, we can cancel it out:
\[ 50 \cdot 19.62 = 30 \cdot \Delta T \]

6. **Solve for ΔT**:
Rearranging the equation to solve for ΔT gives:
\[ \Delta T = \frac{50 \cdot 19.62}{30} \]
Calculating this:
\[ \Delta T = \frac{981}{30} \approx 32.7 \, \text{K} \]

7. **Final approximation**:
The approximate increase in temperature of the mercury is about 32.7 K.