Question 213: Thermodynamics & KTG

Question

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.
(Latent heat of fusion of lead $=2.5 	imes 10^4 \mathrm{JKg}^{-1}$ and specific heat capacity of lead $=125 \mathrm{JKg}^{-1}$ $\left.\mathrm{K}^{-1}ight)$

Accepted Answer

Let's determine the mass step by step:
The bullet is heated from 300 K to 600 K, so the increase in temperature is:
$$\Delta T = 600\, 	ext{K} - 300\, 	ext{K} = 300\, 	ext{K}.$$
The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:
$$Q_1 = m 	imes c 	imes \Delta T = m 	imes 125\, 	ext{J/kg K} 	imes 300\, 	ext{K} = 37500\, m \, 	ext{J},$$
where $ m $ is the mass in kilograms.
After reaching 600 K, the bullet melts. The heat required for the phase change (melting) is given by:
$$Q_2 = m 	imes L = m 	imes 2.5 	imes 10^4\, 	ext{J/kg} = 25000\, m \, 	ext{J}.$$
The total heat required for the process is:
$$Q = Q_1 + Q_2 = 37500\, m + 25000\, m = 62500\, m \, 	ext{J}.$$
We are given that the total heat is 625 J, so:
$$62500\, m = 625.$$
Solving for $ m $:
$$m = \frac{625}{62500} = 0.01\, 	ext{kg}.$$
Finally, converting the mass from kilograms to grams:
$$0.01\, 	ext{kg} = 0.01 	imes 1000\, 	ext{g} = 10\, 	ext{g}.$$
Thus, the mass of the bullet is 10 grams. This corresponds to Option D.

Answer

20

Answer

15

Answer

5

Answer

10

Thermodynamics & KTG

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