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)

A thermally insulated pot has 150 g ice at temperature 0 ""^@C. How mu

1.	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» Solution :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"` Formulai. 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` intowater `0^@C (Q_3)` `m xx L_("melt")` iv. Heat requiredto 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` `therefore590 M = 19500` `therefore M =(19500)/(590) = 33 g `