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6 gm of steam at 100^@ C is mixed with 6...

`6 gm` of steam at `100^@ C` is mixed with `6 gm` of ice at `0^@ C`. Find the mass of steam left uncondensed.
`(L_(f)=80 cal//g,L_(v)= 540cal//g, S_(water) = 1cal//g-^(@) C)`.

Text Solution

Verified by Experts

For steam
Heat lost by the steam in condensation
`Q_(1) = m_(s)L_(s) = 6 xx 540 = 3240 cal` …(1)
For ice
Heat gained by the ice in melting and to rise its temperature from `0^@ C` to `100^@ C` is
`Q_(2) = m_(ice)L_(ice) + m_(ice) S_(w) Delta t`
=`6 xx 80 + 6 xx 1 xx 100 = 1080 cal` ....(2)
From eq (1) and (2) `Q_(1) gt Q_(2)`
i.e., the total steam did not condensed into water.
Let 'm' gm of steam is condensed into water by giving `1080 cal`. of heat.
`mL_(s)=1080 , m = (1080)/(540) = 2gm`
:. mass of the steam left uncondensed `= 6 - 2 = 4g`.
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