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Steam is passed into `54 gm` of water at `30^(@)C` till the temperature of mixture becomes `90^(@)C`. If the latent heat of steam is `536 cal//gm`, the mass of the mixture will beA. `80 gm`B. `60 gm`C. `50 gm`D. `24 gm` |
Answer» Correct Answer - B Let mass of steam condensed = `M gm` `M xx 536 = mxx1xx(100-90)` `=54xx1xx(90-30)` `Mxx536=54xx60` `M = (54xx60)/(536) ~=6gm` Hence, mass of mixture = `54+6 = 60gm`. |
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