Heat extracted to cool x gram of water from 50∘C to 0∘C is given by:
Q1=m1cΔT1
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=m2cΔT2+m2L
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=26762466000≈921.52
The closest integer value for x is 922.
Answer: 922