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, specific heat capacity of water 4200 J/kg. K)
Answer
922
Solution
Heat given out by x g of water cooling from 50 °C to 0 °C:
Q1= x1000 4200 50=210x J
Heat needed by (1000-x) g: warm 50 °C 100 °C, then evaporate:
Q2= 1000-x1000 (4200 50+2256 103 )
=(1000-x)(210+2256)=2466(1000-x) J
Q1=Q2 210x=2466(1000-x)
2676x=2466000 x=921.5
x 922
Answer: 922