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`20 gm` ice at `-10^(@)C` is mixed with `m gm` steam at `100^(@)C`. The minimum value of `m` so that finally all ice and steam converts into water is: `("Use " s_("ice") = 0.5 "cal gm"^(@)C, S_("water") = 1 cal//gm^(@)C, L`) (melting) `= 80 cal// gm` and `L ("vaporization") = 540 cal//gm`)A. `(185)/(27) gm`B. `(135)/(17) gm`C. `(85)/(32) gm`D. `(113)/(17) gm` |
Answer» Correct Answer - C For minimum value of `m`, the final temperature of the mixture must be `0^(@)C` `:. 20xx1/2xx10+20xx80=m 540+m*1*100` `:. M=1700/640 = 85/32 gm`. |
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