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? (Given: Latent heat of melting of ice 80 cal/g, latent heat of vaporisation of water = 540 cal/g, specific heat of water = 1 cal/`g""^@C)`

Answer

33 g of steam of `100^@C` should be mixed.

Explanation

Given: Mass o ice / water (m)=150 g
Change in temperature of water `(DeltaT)`
`=50 ""^@C` ,
We know that: Latent heat of melting of ice `=L_("melt") = 80 "cal/g"`
Specific heat of water `c_w= 1 cal//g^@C`
To find : Mass of steam (M)
flow of heat: Loss of heat:
`underset((100""^@C))"steam"overset(Q_1)rarrunderset((100""^@C))"water"overset(Q_2)rarr underset((50^@C))"water"`
Gain of water
`underset((0^@C))"ice" overset(Q_3)rarr underset((0^2C))"water"overset(Q_3)rarrunderset((50^@C))"water"`
Formula i. Heat to be given off in converting steam into at `100^@C (Q_1)=ML_("vap")`
ii. Heat to be given off in cooling water at `100 ^@C` to `50^@C(Q_2)= Mxx C_w xx Delta T`
iii. Heat required to convert ice at `0^@C` into water at `0^@ C` into water `0^@C (Q_3)`
`m xx L_("melt")`
iv. Heat required to reaise temperature of water from `0^@C "to " 50^@C(Q_4)`
`m= xx c_w xx Delta T`
Calculation: According to princ of heat exchange,
`Q_1 +Q_2=Q_3+Q_4`
From formulae (i) to (iv),
`(Mxx540)+(Mxx1xx50)`
`=(150 xx 80 )+(150 xx 1 xx 50)`
`M(540+50)=12000 +7500`
`therefore 590 M = 19500`
`therefore M =(19500)/(590) = 33 g `