A gun fires a lead bullet of temperature 300K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is ___grams.
(Latent heat of fusion of lead = 2.5 × 104 JKg^–1 and specific heat capacity of lead = 125 JKg^–1 K^–1)

To find the mass of the lead bullet, we need to consider the heat required to:

Raise the temperature of the bullet from 300 K to its melting point (600 K)

Melt the bullet at 600 K

Step 1: Heat Required to Raise the Temperature

The heat required to raise the temperature of the bullet is given by the formula:

$$Q_1 = m c \Delta T$$

where:

$m$ = mass of the bullet (kg)

$c$ = specific heat capacity of lead = 125 J/kg·K

$\Delta T$ = change in temperature = $600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K}$

So,

$$Q_1 = m \times 125 \times 300$$

 $$Q_1 = 37500 m \, \text{J}$$

Step 2: Heat Required to Melt the Bullet

The heat required to melt the bullet at its melting point is given by:

$$Q_2 = m L$$

where:

$L$ = latent heat of fusion of lead = $2.5 \times 10^4 \, \text{J/kg}$

So,

$$Q_2 = m \times 2.5 \times 10^4$$

 $$Q_2 = 2.5 \times 10^4 m \, \text{J}$$

Step 3: Total Heat Required

The total heat required for both processes is given as 625 J, so:

$$Q_1 + Q_2 = 625$$

Substituting the expressions for $Q_1$ and $Q_2$:

$$37500 m + 2.5 \times 10^4 m = 625$$

 $$62500 m = 625$$

Step 4: Solve for the Mass

$$m = \frac{625}{62500}$$

 $$m = 0.01 \, \text{kg} = 10 \, \text{grams}$$

Final Answer:

The mass of the bullet is 10 grams.