Q.47. The heat extracted out of x gram of water initially at 50°C to cool it down to 0°C is sufficie

Q: What is the correct answer for JEE Main Physics Q47?

The correct answer is 922. Given: Mass of water cooled = x gram Initial temperature = $50^\circ C$, Final temperature = $0^\circ C$ Mass of water evaporated = $(1000 - x)$ gram Initial temperature of water for evaporation = $50^\circ C$ Latent heat of vaporization $L_v = 2256$ kJ/kg = $2256 \times 10^3$ J/kg Specific heat capacity of water $c = 4200$ J/kg. K Heat extracted from x gram of water to cool from $50^\circ C$ to

Q: Q.47. The heat extracted out of x gram of water initially at 50°C to cool it down to 0°C is sufficient to evaporate (1000 - x) gram of water also initially at 50°C. The value of x (closest integer)…

The correct answer is 922. Given: Mass of water cooled = x gram Initial temperature = $50^\circ C$, Final temperature = $0^\circ C$ Mass of water evaporated = $(1000 - x)$ gram Initial temperature of water for evaporation = $50^\circ C$ Latent heat of vaporization $L_v = 2256$ kJ/kg = $2256 \times 10^3$ J/kg Specific heat capacity of water $c = 4200$ J/kg. K Heat extracted from x gram of water to cool from $50^\circ C$ to

Question

Q.47. The heat extracted out of x gram of water initially at 50°C to cool it down to 0°C is sufficient to evaporate (1000 - x) gram of water also initially at 50°C. The value of x (closest integer) is . (Take latent heat of water 2256 kJ/kg. K, specific heat capacity of water 4200 J/kg. K)

FirstInTest · Accepted Answer

Given:
Mass of water cooled = x gram
Initial temperature = $50^\circ C$, Final temperature = $0^\circ C$
Mass of water evaporated = $(1000 - x)$ gram
Initial temperature of water for evaporation = $50^\circ C$
Latent heat of vaporization $L_v = 2256$ kJ/kg = $2256 	imes 10^3$ J/kg
Specific heat capacity of water $c = 4200$ J/kg. K

Heat extracted from x gram of water to cool from $50^\circ C$ to $0^\circ C$:
$Q_1 = mc\Delta T = x 	imes 10^{-3} 	ext{ kg} 	imes 4200 	ext{ J/kg.K} 	imes (50 - 0) 	ext{ K} = x 	imes 4200 	imes 50 	imes 10^{-3} 	ext{ J}$.

Heat required to evaporate $(1000 - x)$ gram of water initially at $50^\circ C$:
First, heat to raise temperature from $50^\circ C$ to $100^\circ C$ (boiling point) is needed for evaporation, but the problem states 'evaporate (1000-x) gram of water also initially at 50°C'. This implies the heat extracted is used to evaporate water *at* 50°C, which is unusual. However, typically evaporation happens at boiling point. Let's assume the problem implies that the heat extracted is used to first raise the temperature of $(1000-x)$ gram of water from $50^\circ C$ to $100^\circ C$, and then evaporate it at $100^\circ C$. But the phrasing "evaporate (1000 - x) gram of water also initially at 50°C" is ambiguous. The solution provided uses $Q_2 = (1000-x) 	imes c 	imes (100-50) + (1000-x) 	imes L_v$. Let's follow the solution's interpretation.

Heat required to raise temperature of $(1000 - x)$ gram of water from $50^\circ C$ to $100^\circ C$:
$Q_{2a} = (1000 - x) 	imes 10^{-3} 	ext{ kg} 	imes 4200 	ext{ J/kg.K} 	imes (100 - 50) 	ext{ K} = (1000 - x) 	imes 4200 	imes 50 	imes 10^{-3} 	ext{ J}$.
Heat required to evaporate $(1000 - x)$ gram of water at $100^\circ C$:
$Q_{2b} = (1000 - x) 	imes 10^{-3} 	ext{ kg} 	imes 2256 	imes 10^3 	ext{ J/kg} = (1000 - x) 	imes 2256 	ext{ J}$.

Total heat required for evaporation process $Q_2 = Q_{2a} + Q_{2b}$.
$Q_1 = Q_2$
$x 	imes 4200 	imes 50 	imes 10^{-3} = (1000 - x) 	imes 4200 	imes 50 	imes 10^{-3} + (1000 - x) 	imes 2256$
$x 	imes 210 = (1000 - x) 	imes 210 + (1000 - x) 	imes 2256$
$210x = 210000 - 210x + 2256000 - 2256x$
$210x + 210x + 2256x = 210000 + 2256000$
$2676x = 2466000$
$x = \frac{2466000}{2676} \approx 921.52 	ext{ gram}$.

Closest integer is 922.

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