Physics Circle home Student Login

Home›Chapters›Thermometry, Thermal Expansion and Calorimetry PYQs›Calorimetry, Latent Heat & Mechanical Equivalent of Heat›Question 13

JEE Main 2025 (23 Jan, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2025 (23 Jan, Shift 1)

Q13JEE Main 2025 · 23 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

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 =2.5×104 J kg−1= 2.5 \times 10^{4}\text{ J kg}^{-1} and specific heat capacity of lead =125 J kg−1 K−1)= 125\text{ J kg}^{-1}\text{ K}^{-1}\Big)

Show answer and solution

Answer: (d)

  1. Heat =m(cΔT+L)=m(125×300+2.5×104)=62500 m= m(c\Delta T + L) = m(125 \times 300 + 2.5 \times 10^4) = 62500\,m
  2. 62500 m=625⇒m=0.01 kg=10 g62500\,m = 625 \Rightarrow m = 0.01\text{ kg} = 10\text{ g}
  3. So the answer is (d) 10.
Report a mistake

More questions on Calorimetry, Latent Heat & Mechanical Equivalent of Heat