JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Heat & Thermodynamics

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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 \times 1 0^{4} JKg^{- 1}

and specific heat capacity of lead

= 125 JKg^{- 1} K^{- 1}

)

(A)20
(B)15
(C)10✓ Correct
(D)5

Answer: (C)

Solution

The bullet is heated from 300 K to its melting point 600 K and then melts — both processes draw from the same 625 J of heat.

Q = m s Δ T + m L

625 = m [125 \times (600 - 300) + 2.5 \times 1 0^{4}]

625 = m [37500 + 25000] = m \times 62500

m = \frac{625}{62500} = \frac{1}{100} kg = 10 g

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option

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