JEE Main 2025 Jan 23 Shift 1, Physics Q12: Change of State And Calorimetry

JEE Main2025Jan 23, Shift 1Physics

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 $= 2.5 \times 1 0^{4}$ J kg $^{- 1}$ and specific heat capacity of lead $= 125$ J kg $^{- 1}$ K $^{- 1}$ )

A
$20$
B
$15$
C
$10$
D
$5$

Solution

Heat needed has two parts: warming the bullet from 300 K to its melting point 600 K, then melting it at constant temperature.

$Q = m s Δ T + m L = m (s Δ T + L)$

$625 = m (125 \times 300 + 2.5 \times 1 0^{4}) = m (37500 + 25000) = 62500 m$ .

$m = \frac{625}{62500} = 0.01$ kg $= 10$ g.

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JEE Main 2025 Jan 23 Shift 1, Physics Q12: Change of State And Calorimetry

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