The heat extracted out of gram of water initially at to cool it down to is sufficient to evaporate gram of water also initially at

Question

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) [2026]

Smart Achievers · Accepted Answer

(922)

Given :

$According to question: Q_{1} = Q_{2}$

${(m s Δ T)}_{1} = {(m s Δ T + m L)}_{2}$

$x \times 4.2 \times 50 = (1000 - x) [4.2 \times 50 + 2256]$

$210 x = (1000 - x) [2466]$

$2676 x = 2466 \times 1000$

$x = 921.52 \approx 922$

Solution

Ans.
(922)

Given :

$According to question: Q_{1} = Q_{2}$

${(m s Δ T)}_{1} = {(m s Δ T + m L)}_{2}$

$x \times 4.2 \times 50 = (1000 - x) [4.2 \times 50 + 2256]$

$210 x = (1000 - x) [2466]$

$2676 x = 2466 \times 1000$

$x = 921.52 \approx 922$

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Solution

Ans. (922) Given : According to question: Q1=Q2 (msΔT)1=(msΔT+mL)2 x×4.2×50=(1000-x)[4.2×50+2256] 210x=(1000-x)[2466] 2676x=2466×1000 x=921.52≈922

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Ans.
(922)

Given :

$According to question: Q_{1} = Q_{2}$

${(m s Δ T)}_{1} = {(m s Δ T + m L)}_{2}$

$x \times 4.2 \times 50 = (1000 - x) [4.2 \times 50 + 2256]$

$210 x = (1000 - x) [2466]$

$2676 x = 2466 \times 1000$

$x = 921.52 \approx 922$