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MCQEasyJEE 2025Calorimetry & Change of State

JEE Physics 2025 Question with Solution

A gun fires a lead bullet of temperature 300 K300 \, \text{K} into a wooden block. The bullet having melting temperature of 600 K600 \, \text{K} penetrates into the block and melts down. If the total heat required for the process is 625 J625 \, \text{J}, then the mass of the bullet is _____ grams. Given Data: Latent heat of fusion of lead = 2.5×104 J kg−12.5 \times 10^4 \, \text{J kg}^{-1} and specific heat capacity of lead = 125 J kg−1 K−1125 \, \text{J kg}^{-1} \text{ K}^{-1}.

  • A

    2020

  • B

    1515

  • C

    1010

  • D

    55

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Initial temperature of the bullet is Ti=300 KT_i = 300 \, \text{K}, melting temperature is Tm=600 KT_m = 600 \, \text{K}, total heat required is Q=625 JQ = 625 \, \text{J}, specific heat capacity of lead is c=125 J kg−1 K−1c = 125 \, \text{J kg}^{-1} \text{ K}^{-1}, and latent heat of fusion is L=2.5×104 J kg−1L = 2.5 \times 10^4 \, \text{J kg}^{-1}.

Find: The mass mm of the lead bullet.

The process has two parts:

  1. Heating the bullet from 300 K300 \, \text{K} to 600 K600 \, \text{K}
  2. Melting the bullet at 600 K600 \, \text{K}

Heat required to raise the temperature:

Q1=mc(Tm−Ti)Q_1 = mc(T_m - T_i)

Heat required to melt the bullet:

Q2=mLQ_2 = mL

Therefore, total heat is:

Q=mc(Tm−Ti)+mLQ = mc(T_m - T_i) + mL

Substituting the given values:

625=m×125×(600−300)+m×2.5×104625 = m \times 125 \times (600 - 300) + m \times 2.5 \times 10^4 625=m×(125×300+2.5×104)625 = m \times (125 \times 300 + 2.5 \times 10^4) 625=m×(37500+25000)625 = m \times (37500 + 25000) 625=m×62500625 = m \times 62500

Solving for mm:

m=62562500=1100 kg=0.01 kg=10 gm = \frac{625}{62500} = \frac{1}{100} \, \text{kg} = 0.01 \, \text{kg} = 10 \, \text{g}

Therefore, the mass of the bullet is 10 g10 \, \text{g}. The correct option is C.

Heat Split Approach

Given: The bullet is heated from 300 K300 \, \text{K} to 600 K600 \, \text{K} and then melts. Total heat supplied is 625 J625 \, \text{J}.

Find: Mass of the bullet.

First calculate the heat needed for temperature rise:

Q1=msΔT=m×125×(600−300)Q_1 = ms\Delta T = m \times 125 \times (600 - 300) Q1=m×125×300Q_1 = m \times 125 \times 300

Now calculate the heat needed for phase change:

Q2=mL=m×2.5×104Q_2 = mL = m \times 2.5 \times 10^4

Using total heat:

625=Q1+Q2625 = Q_1 + Q_2 625=m×125×300+m×2.5×104625 = m \times 125 \times 300 + m \times 2.5 \times 10^4 625=m×37500+m×25000625 = m \times 37500 + m \times 25000 625=m×62500625 = m \times 62500

Hence,

m=62562500=1100 kgm = \frac{625}{62500} = \frac{1}{100} \, \text{kg}

Converting to grams:

m=10 gm = 10 \, \text{g}

Therefore, the mass of the bullet is 10 g10 \, \text{g}, so the correct option is C.

Common mistakes

  • Adding only the latent heat term and ignoring the heat needed to raise the bullet from 300 K300 \, \text{K} to 600 K600 \, \text{K} is incorrect because the bullet must first reach its melting point. Use both mcΔTmc\Delta T and mLmL.

  • Using the temperature change as 600600 instead of 600−300=300 K600 - 300 = 300 \, \text{K} is wrong because heat capacity uses the change in temperature, not the final temperature. Always compute ΔT\Delta T first.

  • Forgetting that the mass obtained from the equation is in kilograms leads to a wrong final value. After finding m=0.01 kgm = 0.01 \, \text{kg}, convert it to grams to get 10 g10 \, \text{g}.

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