Question
An iron nail is driven into a block of ice by a single blow of a hammer. The hammerhead has a mass of 0.3 kg and an initial speed of 1.9 m/s. Nail and hammer are at rest after the blow. How much ice melts? Assume the temperature of both the ice and the nail is 0◦C before and after. The heat of fusion of ice is 80 cal/g. Answer in units of g.
Answers
GPT-4o mini
To find out how much ice melts when the iron nail is driven into the block of ice, we can start by calculating the kinetic energy of the hammer before it strikes the nail.
The kinetic energy \( KE \) of the hammer can be calculated using the formula:
\[
KE = \frac{1}{2} mv^2
\]
where \( m \) is the mass of the hammer and \( v \) is its speed. Given that:
- \( m = 0.3 \, \text{kg} \)
- \( v = 1.9 \, \text{m/s} \)
Substituting the values:
\[
KE = \frac{1}{2} \times 0.3 \, \text{kg} \times (1.9 \, \text{m/s})^2
\]
Calculating \( (1.9 \, \text{m/s})^2 \):
\[
(1.9)^2 = 3.61 \, \text{m}^2/\text{s}^2
\]
Now plug this back into the kinetic energy formula:
\[
KE = \frac{1}{2} \times 0.3 \, \text{kg} \times 3.61 \, \text{m}^2/\text{s}^2
\]
\[
KE = 0.15 \, \text{kg} \times 3.61 \, \text{m}^2/\text{s}^2
\]
\[
KE = 0.5415 \, \text{kg m}^2/\text{s}^2 = 0.5415 \, \text{J}
\]
Now, we can convert this energy into calories, since the heat of fusion of ice is given in calories. The conversion factor is:
\[
1 \, \text{cal} = 4.184 \, \text{J}
\]
Thus, to convert kinetic energy from joules to calories:
\[
KE_{\text{cal}} = \frac{0.5415 \, \text{J}}{4.184 \, \text{J/cal}} \approx 0.129 \, \text{cal}
\]
Next, we need to find out how much ice can be melted with this amount of energy. The heat of fusion of ice is given as 80 cal/g. We can find the mass of the ice melted (\( m \)) using the formula:
\[
Q = m \cdot L_f
\]
Where:
- \( Q \) = heat energy (in calories)
- \( L_f \) = heat of fusion (80 cal/g)
Rearranging for \( m \):
\[
m = \frac{Q}{L_f}
\]
Substituting the values:
\[
m = \frac{0.129 \, \text{cal}}{80 \, \text{cal/g}} \approx 0.0016125 \, \text{g}
\]
Thus, the amount of ice that melts when the hammer strikes the iron nail is approximately:
\[
m \approx 0.0016125 \, \text{g} \approx 0.0016 \, \text{g}
\]
In conclusion, the amount of ice melted is approximately:
\[
\boxed{0.0016 \, \text{g}}
\]
Note: Realistically, this value is much smaller than anticipated, implying that very little ice would melt from the energy transferred in this scenario.
The kinetic energy \( KE \) of the hammer can be calculated using the formula:
\[
KE = \frac{1}{2} mv^2
\]
where \( m \) is the mass of the hammer and \( v \) is its speed. Given that:
- \( m = 0.3 \, \text{kg} \)
- \( v = 1.9 \, \text{m/s} \)
Substituting the values:
\[
KE = \frac{1}{2} \times 0.3 \, \text{kg} \times (1.9 \, \text{m/s})^2
\]
Calculating \( (1.9 \, \text{m/s})^2 \):
\[
(1.9)^2 = 3.61 \, \text{m}^2/\text{s}^2
\]
Now plug this back into the kinetic energy formula:
\[
KE = \frac{1}{2} \times 0.3 \, \text{kg} \times 3.61 \, \text{m}^2/\text{s}^2
\]
\[
KE = 0.15 \, \text{kg} \times 3.61 \, \text{m}^2/\text{s}^2
\]
\[
KE = 0.5415 \, \text{kg m}^2/\text{s}^2 = 0.5415 \, \text{J}
\]
Now, we can convert this energy into calories, since the heat of fusion of ice is given in calories. The conversion factor is:
\[
1 \, \text{cal} = 4.184 \, \text{J}
\]
Thus, to convert kinetic energy from joules to calories:
\[
KE_{\text{cal}} = \frac{0.5415 \, \text{J}}{4.184 \, \text{J/cal}} \approx 0.129 \, \text{cal}
\]
Next, we need to find out how much ice can be melted with this amount of energy. The heat of fusion of ice is given as 80 cal/g. We can find the mass of the ice melted (\( m \)) using the formula:
\[
Q = m \cdot L_f
\]
Where:
- \( Q \) = heat energy (in calories)
- \( L_f \) = heat of fusion (80 cal/g)
Rearranging for \( m \):
\[
m = \frac{Q}{L_f}
\]
Substituting the values:
\[
m = \frac{0.129 \, \text{cal}}{80 \, \text{cal/g}} \approx 0.0016125 \, \text{g}
\]
Thus, the amount of ice that melts when the hammer strikes the iron nail is approximately:
\[
m \approx 0.0016125 \, \text{g} \approx 0.0016 \, \text{g}
\]
In conclusion, the amount of ice melted is approximately:
\[
\boxed{0.0016 \, \text{g}}
\]
Note: Realistically, this value is much smaller than anticipated, implying that very little ice would melt from the energy transferred in this scenario.