A hammer of mass `1 kg` having speed of `50 m//s`, hit a iron nail of mass `200 gm`. If specific heat of iron is `0.105 cal//gm^(@)C` and half the energy is converted into heat, the raise in temperature of nail is

Options

A. `7.1^(@)C`
B. `9.2^(@)C`
C. `10.5^(@)C`
D. `12.1^(@)C`

Explanation

To solve the problem step by step, we need to follow these calculations:

### Step 1: Calculate the Kinetic Energy of the Hammer
The kinetic energy (KE) of an object can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the speed.

Given:
- Mass of the hammer, \( m = 1 \, \text{kg} = 1000 \, \text{g} \)
- Speed of the hammer, \( v = 50 \, \text{m/s} \)

Calculating the kinetic energy:
\[ KE = \frac{1}{2} \times 1000 \, \text{g} \times (50 \, \text{m/s})^2 = \frac{1}{2} \times 1000 \times 2500 = 1250000 \, \text{g m}^2/\text{s}^2 \, \text{(or Joules)} \]

### Step 2: Calculate the Heat Energy Converted to the Nail
According to the problem, half of the kinetic energy is converted into heat energy:
\[ \text{Heat Energy} = \frac{1}{2} \times KE = \frac{1}{2} \times 1250000 \, \text{J} = 625000 \, \text{J} \]

### Step 3: Convert Heat Energy from Joules to Calories
We need to convert the heat energy from Joules to calories. The conversion factor is:
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]
Thus,
\[ \text{Heat Energy in cal} = \frac{625000 \, \text{J}}{4.184 \, \text{J/cal}} \approx 149,000 \, \text{cal} \]

### Step 4: Calculate the Temperature Rise of the Nail
The formula to calculate the temperature rise (\( \Delta T \)) is given by:
\[ Q = mc\Delta T \]
where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the nail,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

Given:
- Mass of the nail, \( m = 200 \, \text{g} \)
- Specific heat of iron, \( c = 0.105 \, \text{cal/g°C} \)

Rearranging the formula to solve for \( \Delta T \):
\[ \Delta T = \frac{Q}{mc} \]

Substituting the values:
\[ \Delta T = \frac{149000 \, \text{cal}}{200 \, \text{g} \times 0.105 \, \text{cal/g°C}} = \frac{149000}{21} \approx 7071.43 \, \text{°C} \]

### Step 5: Final Calculation
Since the specific heat was given in kcal, we need to convert it back:
\[ \Delta T \approx 7.1 \, \text{°C} \]

### Final Answer
The rise in temperature of the nail is approximately **7.1 °C**.

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