The 0.50 kg head of a hammer has speed of 5.0 m//s just before its strikes a nail and is brought to rest. Estimate the temperature rise a 15 g nail generated by ten such hammer blows done in quick succession. Assume the nail absorbs all the heat.

Answer

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

### Step 1: Calculate the change in kinetic energy of the hammer
The kinetic energy (KE) of the hammer just before it strikes the nail can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where:
- \( m = 0.50 \, \text{kg} \) (mass of the hammer)
- \( v = 5.0 \, \text{m/s} \) (speed of the hammer)

Substituting the values:
\[ KE = \frac{1}{2} \times 0.50 \times (5.0)^2 = \frac{1}{2} \times 0.50 \times 25 = 6.25 \, \text{J} \]

### Step 2: Calculate the total energy transferred to the nail after 10 blows
Since the hammer strikes the nail 10 times, the total energy transferred to the nail will be:
\[ \text{Total Energy} = \text{Energy per blow} \times \text{Number of blows} = 6.25 \, \text{J} \times 10 = 62.5 \, \text{J} \]

### Step 3: Use the heat transfer equation to find the temperature rise
The heat generated (which is absorbed by the nail) can be calculated using the formula:
\[ Q = mc\Delta T \]
where:
- \( Q = 62.5 \, \text{J} \) (total heat energy absorbed by the nail)
- \( m = 15 \, \text{g} = 0.015 \, \text{kg} \) (mass of the nail)
- \( c = 0.444 \, \text{J/(g°C)} \) (specific heat capacity of iron)

Rearranging the formula to find the temperature rise (\( \Delta T \)):
\[ \Delta T = \frac{Q}{mc} \]

Substituting the values:
\[ \Delta T = \frac{62.5}{0.015 \times 0.444} \]

Calculating the denominator:
\[ 0.015 \times 0.444 = 0.00666 \, \text{kg°C} \]

Now substituting this back into the equation:
\[ \Delta T = \frac{62.5}{0.00666} \approx 9403.6 \, °C \]

### Step 4: Conclusion
The temperature rise of the nail after 10 hammer blows is approximately:
\[ \Delta T \approx 9403.6 \, °C \]