Physics · Thermal Properties of Matter

JEE Main 2025 — 23 January, Morning Shift — Question 59

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 $\qquad$ grams.

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

Option A:
20
Option B:
15
Option C:
10
Correct
Option D:
5

Answer: C

Step-by-step solution

$625=\mathrm{ms} \Delta \mathrm{T}+\mathrm{mL}$

$625=\mathrm{m}\left[125 \times 300+2.5 \times 10^{4}\right]$

$625=\mathrm{m}[37500+25000]$

$625=\mathrm{m}[62500]$

$\mathrm{m}=\frac{1}{100} \mathrm{~kg}$

$\mathrm{M}=10$ grams

Answer key and solution verified before publishing.

Practise Thermal Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermometry and Calorimetry