1 g of steam at 100° C and equal mass of ice at 0° C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 text cal/g, latent heat of ice = 80 cal/g ):

Question

1 g of steam at 100° C and equal mass of ice at 0° C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 text cal/g, latent heat of ice = 80 cal/g ): (A)  50° C  (B)  100° C  (C)  67° C  (D)  33° C

Tardigrade Admin · Accepted Answer

Heat taken by ice to raise its temperature, to 100° C Q1=1× 80+1× 1× 100 =180 text cal Heat given by steam when condensed Q2=m2L2 =1× 540=540 text cal As Q2>Q1, hence, temperature of mixture will remain 100° C .

Q. 1 g of steam at $100^{\circ} C$ and equal mass of ice at $0^{\circ} C$ are mixed. The temperature of the mixture in steady state will be (latent heat of steam $= 540 c a l / g,$ latent heat of ice = 80 $c a l / g$ ):

$50^{\circ} C$

$100^{\circ} C$

$67^{\circ} C$

$33^{\circ} C$

$0^{\circ} C$

Heat taken by ice to raise its temperature, to $100^{\circ} C$ $Q_{1} = 1 \times 80 + 1 \times 1 \times 100$ $= 180 c a l$ Heat given by steam when condensed $Q_{2} = m_{2} L_{2}$ $= 1 \times 540 = 540 c a l$ As $Q_{2} > Q_{1},$ hence, temperature of mixture will remain $100^{\circ} C$ .

Q. 1 g of steam at 100∘C and equal mass of ice at 0∘C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540cal/g, latent heat of ice = 80 cal/g ):

50∘C

100∘C

67∘C

33∘C

0∘C

Heat taken by ice to raise its temperature, to 100∘C Q1​=1×80+1×1×100 =180cal Heat given by steam when condensed Q2​=m2​L2​ =1×540=540cal As Q2​>Q1​, hence, temperature of mixture will remain 100∘C .