JEE Challenger
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PhyMedMain 2026 • 5th Apr - Shift 2

Heat Extracted from Water to Evaporate Remaining Water

The heat extracted out of x gramx\text{ gram} of water initially at 50 ∘C50\ ^\circ\text{C} to cool it down to 0 ∘C0\ ^\circ\text{C} is sufficient to evaporate (1000−x) gram(1000 - x)\text{ gram} of water also initially at 50 ∘C50\ ^\circ\text{C}. The value of xx (closest integer) is ______.
(Take latent heat of water 2256 kJ/kg⋅K2256\text{ kJ/kg}\cdot\text{K}, specific heat capacity of water 4200 J/kg⋅K4200\text{ J/kg}\cdot\text{K})

Official Numerical Answer922

Topics & Concepts

Thermal Properties of MatterCalorimetryHeat Transfer

Step-by-Step Solution

To find the value of xx, we calculate the total heat extracted from the first portion of water and equate it to the total heat required to evaporate the remaining portion of water.

1. Heat Extracted to Cool x gx\text{ g} of Water

The amount of heat extracted to cool x gx\text{ g} of water from 50 ∘C50\ ^\circ\text{C} to 0 ∘C0\ ^\circ\text{C} is given by: Q1=m1⋅c⋅ΔT1Q_1 = m_1 \cdot c \cdot \Delta T_1

Given:

  • Mass m1=x g=x×10−3 kgm_1 = x\text{ g} = x \times 10^{-3}\text{ kg}
  • Specific heat capacity c=4200 J/kg⋅K=4.2 J/g⋅∘Cc = 4200\text{ J/kg}\cdot\text{K} = 4.2\text{ J/g}\cdot^\circ\text{C}
  • Change in temperature ΔT1=50 ∘C−0 ∘C=50 ∘C\Delta T_1 = 50\ ^\circ\text{C} - 0\ ^\circ\text{C} = 50\ ^\circ\text{C}

Q1=x×4.2×50=210x JQ_1 = x \times 4.2 \times 50 = 210 x\text{ J}

2. Heat Required to Evaporate (1000−x) g(1000 - x)\text{ g} of Water

To evaporate water initially at 50 ∘C50\ ^\circ\text{C}, it must first be heated to its boiling point (100 ∘C100\ ^\circ\text{C}) and then vaporized into steam at 100 ∘C100\ ^\circ\text{C}.

Given:

  • Mass m2=(1000−x) gm_2 = (1000 - x)\text{ g}
  • Latent heat of vaporization L=2256 kJ/kg=2256 J/gL = 2256\text{ kJ/kg} = 2256\text{ J/g}

Step A: Heat required to raise temperature from 50 ∘C50\ ^\circ\text{C} to 100 ∘C100\ ^\circ\text{C} Qsensible=m2⋅c⋅ΔT2Q_{\text{sensible}} = m_2 \cdot c \cdot \Delta T_2 Qsensible=(1000−x)×4.2×(100−50)=210(1000−x) JQ_{\text{sensible}} = (1000 - x) \times 4.2 \times (100 - 50) = 210(1000 - x)\text{ J}

Step B: Heat required for vaporization at 100 ∘C100\ ^\circ\text{C} Qlatent=m2⋅LQ_{\text{latent}} = m_2 \cdot L Qlatent=(1000−x)×2256 JQ_{\text{latent}} = (1000 - x) \times 2256\text{ J}

Total Heat Required (Q2Q_2): Q2=Qsensible+QlatentQ_2 = Q_{\text{sensible}} + Q_{\text{latent}} Q2=(1000−x)(210+2256)=2466(1000−x) JQ_2 = (1000 - x)(210 + 2256) = 2466(1000 - x)\text{ J}

3. Equating the Heat Quantities

According to the problem, Q1=Q2Q_1 = Q_2: 210x=2466(1000−x)210 x = 2466(1000 - x) 210x=2466000−2466x210 x = 2466000 - 2466 x 2676x=24660002676 x = 2466000 x=24660002676≈921.52x = \frac{2466000}{2676} \approx 921.52

Rounding off to the closest integer gives: x≈922x \approx 922

Heat Extracted from Water to Evaporate Remaining Water | Physics PYQ Solution - JEE Challenger