A gun fires a lead bullet of temperature 300 K 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. Given Data: Latent heat of fusion of lead = \(2.5 \times 10^4 \, \text{J kg}^{-1}\) and specific heat capacity of lead = 125 J kg\(^{-1}\) K\(^{-1}\).
A gun fires a lead bullet of temperature 300 K 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. Given Data: Latent heat of fusion of lead = \(2.5 \times 10^4 \, \text{J kg}^{-1}\) and specific heat capacity of lead = 125 J kg\(^{-1}\) K\(^{-1}\).
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- 20
- 15
- 10
- 5
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to calculate the mass of the bullet that melts upon heating. We are given the following data:
- Initial temperature of the bullet, \(T_i = 300 \text{ K}\)
- Melting temperature of the bullet, \(T_m = 600 \text{ K}\)
- Total heat required, \(Q = 625 \text{ J}\)
- Specific heat capacity of lead, \(c = 125 \text{ J kg}^{-1} \text{ K}^{-1}\)
- Latent heat of fusion of lead, \(L = 2.5 \times 10^4 \text{ J kg}^{-1}\)
The problem involves two main heat processes:
- Heating the bullet from \(300 \text{ K}\) to \(600 \text{ K}\).
- Melting the bullet at \(600 \text{ K}\).
The total heat required can be divided into two parts:
- Heat required to raise the temperature: \(Q_1 = mc (T_m - T_i)\)
- Heat required to melt the bullet: \(Q_2 = mL\)
The total heat \(Q\) is given by the sum of \(Q_1\) and \(Q_2\):
\(Q = mc (T_m - T_i) + mL\)
Substituting the known values:
\(625 = m \times 125 \times (600 - 300) + m \times 2.5 \times 10^4\)
Simplifying this equation:
\(625 = m \times (125 \times 300 + 2.5 \times 10^4)\)
\(625 = m \times (37500 + 25000)\)
\(625 = m \times 62500\)
Solving for \(m\):
\(m = \frac{625}{62500} = \frac{1}{100} \ \text{kg} = 0.01 \ \text{kg} = 10 \ \text{g}\)
Thus, the mass of the bullet is 10 grams.
Approach Solution -2
The total heat required is the sum of the heat needed to raise the temperature of the bullet from 300 K to 600 K and the heat required to melt the bullet at the melting point.
Using the formula for heat \( Q = ms \Delta T \), where:
\( m \) is the mass,
\( s \) is the specific heat,
\( \Delta T \) is the change in temperature.
The first part of the heat is required to raise the temperature: \[ Q_1 = ms\Delta T = m \times 125 \times (600 - 300) = m \times 125 \times 300 \] The second part is required to melt the bullet at 600 K: \[ Q_2 = mL = m \times (2.5 \times 10^4) \] The total heat is given as 625 J: \[ 625 = ms\Delta T + mL \] Substitute the values: \[ 625 = m \times 125 \times 300 + m \times 2.5 \times 10^4 \] Simplifying: \[ 625 = m \times 37500 + m \times 25000 \] \[ 625 = m \times 62500 \] Solving for \( m \): \[ m = \frac{625}{62500} = \frac{1}{100} \, \text{kg} \] Since \( 1 \, \text{kg} = 1000 \, \text{grams}, \) we have: \[ m = 10 \, \text{grams} \] Thus, the mass of the bullet is 10 grams.
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