A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absor

Q: Why does the mass of the bullet not matter?

Because both the kinetic energy and the heat required to raise the temperature and melt the bullet are directly proportional to its mass. When you equate them, the mass cancels out from both sides of the equation.

Q: Why do we multiply by 1000 when converting units?

The given units are per gram (e.g., cal/g). Since the standard SI unit for mass is the kilogram (kg), and 1 kg = 1000 g, we multiply by 1000 to find the energy required per kilogram.

Q: What happens to the other 25% of the heat?

As stated in the problem, it is absorbed by the obstacle that stops the bullet. It contributes to heating up the obstacle or causing local deformation.

Q: Why do we add $ms\Delta\theta$ and $mL$?

Melting is a two-step process here. First, the solid lead must be heated from its initial temperature ($27^\circ\text{C}$) to its melting point ($327^\circ\text{C}$). This requires sensible heat ($ms\Delta\theta$). Then, it needs additional energy to actually change state from solid to liquid at that constant temperature. This is the latent heat ($mL$).

Q: What if the calculated velocity was lower?

If the velocity was lower, the kinetic energy would be less. The bullet would heat up, but it might not reach the melting point, or it might only partially melt.

Question

A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27^\circ	ext{C}$.
(Melting point of lead = $327^\circ	ext{C}$, specific heat of lead = $0.03 	ext{ cal/g } ^\circ	ext{C}$, latent heat of fusion of lead = $6 	ext{ cal/g}$, $J = 4.2 	ext{ J/cal}$.)

Sigma Percentile · Accepted Answer

### The High-Speed Collision

Imagine a lead bullet flying through the air at a high velocity $v$. Suddenly, it strikes an obstacle and comes to a complete halt.

What happens to all that kinetic energy? It doesn't just vanish. According to the law of conservation of energy, it transforms into heat.

### The Energy Distribution

The total kinetic energy of the bullet is given by the familiar formula $\frac{1}{2}mv^2$.

However, not all of this energy stays with the bullet. The problem states that the obstacle absorbs $25\%$ of the heat generated upon impact.

This means the bullet retains the remaining $75\%$ of the heat. We can express this retained heat as:
$$Q_{	ext{retained}} = 0.75 	imes \frac{1}{2}mv^2 = \frac{3}{8}mv^2$$

### The Thermodynamics of Melting

Now, what does this retained heat do to the bullet? It must accomplish two things to completely melt it.

First, it must raise the bullet's temperature from its initial state of $27^\circ	ext{C}$ to its melting point of $327^\circ	ext{C}$. This requires sensible heat, calculated as $ms\Delta	heta$.

Second, once at the melting point, it must provide the latent heat required to change the lead from a solid to a liquid state. This is calculated as $mL$.

So, the total heat required is:
$$Q_{	ext{req}} = ms\Delta	heta + mL$$

### The Crucial Unit Conversion

Before we equate the retained heat to the required heat, we must address a major trap: **units**.

The specific heat $s$ and latent heat $L$ are given in calories per gram. We must convert them to standard SI units (Joules per kilogram) to match the kinetic energy which is in Joules.

For specific heat:
$$s = 0.03 	ext{ cal/g}^\circ	ext{C} = 0.03 	imes 4.2 	imes 1000 	ext{ J/kg}^\circ	ext{C} = 126 	ext{ J/kg}^\circ	ext{C}$$

For latent heat:
$$L = 6 	ext{ cal/g} = 6 	imes 4.2 	imes 1000 	ext{ J/kg} = 25200 	ext{ J/kg}$$

### The Final Calculation

Now we can safely equate the energies. Notice how the mass $m$ beautifully cancels out from both sides!
$$\frac{3}{8}v^2 = s\Delta	heta + L$$

Substituting our SI values:
$$\frac{3}{8}v^2 = 126 	imes (327 - 27) + 25200$$

$$\frac{3}{8}v^2 = 126 	imes 300 + 25200$$

$$\frac{3}{8}v^2 = 37800 + 25200 = 63000$$

Solving for $v^2$:
$$v^2 = 63000 	imes \frac{8}{3} = 168000$$

Taking the square root gives us our final velocity:
$$v = \sqrt{168000} \approx \mathbf{409.8 	ext{ m/s}}$$

This is a fantastic problem that elegantly combines mechanics and thermodynamics, reminding us to always keep a watchful eye on our units!

Enter Numerical Value:

Visualized Solution

$Visualizing the Collision$

$Energy Distribution$

$Thermodynamics of Melting$

$Equating Energies$

$Crucial Unit Conversion$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Calorimetry

The High-Speed Collision

The Energy Distribution

The Thermodynamics of Melting

The Crucial Unit Conversion

The Final Calculation

Similar Questions

A substance of mass $M kg$ requires a power input of $P watts$ to remain in the molten state at its melting point. When the power source is turned off, the sample completely solidifies in time $t seconds$ . The latent heat of fusion of the substance is …… .

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Distribution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermodynamics of Melting

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Crucial Unit Conversion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward