Thermal Properties of Matter

JEE Main level, numerical answer, medium

48 of 126

Question 48 of 126

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

Show answer and solution

Answer: 922

Heat extracted to cool xx gram of water from 50∘C50^{\circ}\text{C} to 0∘C0^{\circ}\text{C} is given by:

Q1=m1cΔT1Q_1 = m_1 c \Delta T_1

Q1=x×4.2×(50−0)=210x JQ_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}

Heat required to raise the temperature of (1000−x)(1000 - x) gram of water from 50∘C50^{\circ}\text{C} to 100∘C100^{\circ}\text{C} and evaporate it is given by:

Q2=m2cΔT2+m2LQ_2 = m_2 c \Delta T_2 + m_2 L

Q2=(1000−x)×[4.2×(100−50)+2256]Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]

Q2=(1000−x)×(210+2256)=2466(1000−x) JQ_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}

Equating the heat extracted and the heat required:

Q1=Q2Q_1 = Q_2

210x=2466(1000−x)210x = 2466(1000 - x)

210x=2466000−2466x210x = 2466000 - 2466x

2676x=24660002676x = 2466000

x=24660002676≈921.52x = \dfrac{2466000}{2676} \approx 921.52

The closest integer value for xx is 922922.

Answer: 922922

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