Steam at `100^(@)C` is added slowly to 1400 gm of water at `16^(@)C` until the temperature of water is raised to `80^(@)C`. The mass of steam required
Steam at `100^(@)C` is added slowly to 1400 gm of water at `16^(@)C` until the temperature of water is raised to `80^(@)C`. The mass of steam required to do this is (`L_(V)`= 540 cal/g)
A. 120 gm
B. 240 gm
C. 160 gm
D. 320 gm
5 views
1 Answers
Correct Answer - C
Heat gain by water = Heat loss by steam
`rArr MS_(w)(80-16)=mL.+ms(100-80)`
`rArr 1400 xx 1 xx 64=m xx 540 +m xx 1 xx 20`
`rArr m = 160 ` gram
5 views
Answered