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 \underline\hspace2cm grams. (Latent heat of fusion of lead =2.5 \times 10^4 JKg^-1 and specific heat capacity of lead =125 JKg^-1.K^-1)

Heat and Thermodynamics PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q57 with Solution

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Correct answer: D

The lead bullet is initially at $300\,\text{K}$ and its melting point is $600\,\text{K}$ .

So, the temperature rise is

\Delta T = 600-300 = 300\,\text{K}

Heat required to raise temperature of mass $m$ is

Q_1 = mc\Delta T

Given:

c = 125\,\text{J kg}^{-1}\text{K}^{-1}

Thus,

Q_1 = m \times 125 \times 300 = 37500m

Latent heat of fusion of lead:

L = 2.5\times 10^4\,\text{J kg}^{-1}

So,

Q_2 = mL = m(2.5\times 10^4) = 25000m

Since the bullet is heated to melting point and then completely melts,

Q = Q_1 + Q_2

Given total heat required is $625\,\text{J}$ , so

625 = 37500m + 25000m

625 = 62500m

m = \frac{625}{62500} = 0.01\,\text{kg}

0.01\,\text{kg} = 10\,\text{g}

So the correct option is

\boxed{\text{D: }10\text{ g}}