A mass m of lead shot is placed at the bottom of a vertical cardboard cylinder that is 1.5 m long and closed at both ends. The cylinder is suddenly inverted so that the shot falls 1.5 m If this procces is repeated quickely 100 times, auuming no heat is dissipated or lost, the temperature of the shot will increase by (specific heat of lead`=0.03 cal//g^@C`)

Options

A. 0
B. `5^@C`
C. `7.3^@C`
D. `11.3^@C`

Explanation

To solve the problem step by step, we will follow the principles of energy conservation and heat transfer.

### Step 1: Understand the Problem
We have a mass \( m \) of lead shot that falls a distance of \( h = 1.5 \, \text{m} \) when the cylinder is inverted. This process is repeated \( 100 \) times. We need to find the increase in temperature \( \Delta T \) of the lead shot after these repetitions.

### Step 2: Calculate the Potential Energy Lost
When the lead shot falls, it loses potential energy, which is converted into heat energy, increasing the temperature of the shot. The potential energy lost when the shot falls a distance \( h \) is given by:
\[ PE = mgh \]
where:
- \( m \) = mass of the lead shot (in kg)
- \( g \) = acceleration due to gravity \( \approx 9.8 \, \text{m/s}^2 \)
- \( h = 1.5 \, \text{m} \)

### Step 3: Calculate Total Energy After 100 Falls
Since the shot falls \( 100 \) times, the total potential energy lost is:
\[ \text{Total PE} = 100 \times mgh \]

### Step 4: Relate Energy to Temperature Change
The energy lost by the lead shot is used to increase its temperature. The heat gained by the shot can be expressed as:
\[ Q = mc\Delta T \]
where:
- \( c = 0.03 \, \text{cal/g}^\circ C \) (specific heat of lead)
- \( \Delta T \) = temperature change in degrees Celsius

### Step 5: Set Up the Equation
Setting the total potential energy equal to the heat gained, we have:
\[ 100 \times mgh = mc\Delta T \]

### Step 6: Cancel Mass from Both Sides
Since \( m \) appears on both sides of the equation, we can cancel it:
\[ 100gh = c\Delta T \]

### Step 7: Solve for Temperature Change
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{100gh}{c} \]

### Step 8: Substitute Values
Now, substituting the known values:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 1.5 \, \text{m} \)
- Convert \( c \) from cal/g°C to J/kg°C:
- \( 0.03 \, \text{cal/g}^\circ C = 0.03 \times 1000 \, \text{cal/kg}^\circ C = 30 \, \text{cal/kg}^\circ C \)
- \( 1 \, \text{cal} = 4.18 \, \text{J} \), thus \( 30 \, \text{cal/kg}^\circ C = 30 \times 4.18 \, \text{J/kg}^\circ C = 125.4 \, \text{J/kg}^\circ C \)

Now substituting these values into the equation:
\[ \Delta T = \frac{100 \times 9.8 \times 1.5}{125.4} \]

### Step 9: Calculate \( \Delta T \)
Calculating the above expression:
\[ \Delta T = \frac{100 \times 9.8 \times 1.5}{125.4} \approx \frac{1470}{125.4} \approx 11.7 \, \text{°C} \]

### Final Answer
The temperature of the lead shot will increase by approximately \( 11.7 \, \text{°C} \).

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