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 )

Step-by-step solution

Heat extracted to cool x gram of water from 50^ C to 0^ C is given by: Q₁ = m₁ c T₁ Q₁ = x 4.2 (50 - 0) = 210x J Heat required to raise the temperature of (1000 - x) gram of water from 50^ C to 100^ C and evaporate it is given by: Q₂ = m₂ c T₂ + m₂ L Q₂ = (1000 - x) [4.2 (100 - 50) + 2256] Q₂ = (1000 - x) (210 + 2256) = 2466(1000 - x) J Equating the heat extracted and the heat required: Q₁ = Q₂ 210x = 2466(1000 - x) 210x = 2466000 - 2466x 2676x = 2466000 x = 2466000 2676 921.52 The closest integer value for x is 92