A thermally insulated pot has 150 g ice at temperature ` 0^(@)C` . How much steam of `100^(@)C` has to be mixed to it, so that water of temperature `50^(@)C` will be obtained ?

Explanation

To find : `m_(1) = ?`
Formulae : `Q_(A) = Q_(B)`
`Q = mcDelta, Q = mL`
Solution :
(1) Steam
`"Steam"(100^(@)C) overset(mL)to "Water"(100^(@)C)overset(mcDeltaT)to "Water"(50^(@)C)`
`therefore Q_(A) = m_(1),L_(1) + m_(1)c_(1)DeltaT_(1)............(i)`
(2) Ice
`"Ice"(0^(@)C)overset(mL)to"Water"(0^(@)C)overset(mcDeltaT)to"Water"(50^(@)C)`
`therefore Q_(B) = mL + mcDeltaT.............(ii)`
form (i) and(ii)
`Q_(B) = Q_(A)`
`mL + mcDeltaT = m_(1)L_(1) + m_(1)c_(1)DeltaT`
`(150 xx 80) + {150 xx 1 xx 50 -0)} = (m_(1) xx 540} + {m_(1) xx 1 xx (100 -50)}`
`12000+ 7500 = m_(1) (540 + 50)`
`19500= m_(1)(540 + 50)`
`m_(1) = (1950cancel0)/(590cancel0)`
`m_(1) = 33.05 = 33`