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`50 g` of steam at `100^@ C` is passed into `250 g` of at `0^@ C`. Find the resultant temperature (if latent heat of steam is `540 cal//g`, latent heat of ice is `80 cal//g` and specific heat of water is `1 cal//g -.^@ C`).A. `40^@ C`B. `30^@ C`C. `20^@ C`D. `10^@ C` |
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Answer» Correct Answer - A Heat lost by steam = Heat gained by ice. `m_(steam) L_(v)+m_(s) xxS_(w)(100^@-theta^@)` `=m_(w)L_(f) +m_(w)S_(w) (theta^@ - 0)`. |
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