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Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) isA. `27.33 g`B. `24.8 g`C. `2.8 g`D. `30 g` |
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Answer» Correct Answer - B `m_(steam)L_(v) +m_(steam) S_(w) (100^@ -90^@)` =`m_(w)S_(w)(90^@-20^@)` `m_(steam)` = mass f steam converted into water :. Mass of water `= 22 g + m_(steam)`. |
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