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Latent Heat

One gram of i...

Question

One $gram$ of ice is mixed with one $gram$ of steam. After thermal equilibrium is reached, the temperature of the mixture is
(Latent heat of fusion of ice; $L_{i c e} = 80 cal/g$ , latent heat of condensation of vapour; $L_{V} = 540 cal/g$ ,
Specific heat capacity of water; $c = 1 {cal/g}^{\circ} C$ )

100^{\circ} C

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55^{\circ} C

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50^{\circ} C

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0^{\circ} C

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Solution

The correct option is A $100^{\circ} C$
We know,
Latent heat of fusion of ice; $L_{i c e} = 80 cal/g$
Latent heat of condensation of vapour; $L_{V} = 540 cal/g$
Specific heat capacity of water; $c = 1 {cal/g}^{\circ} C$
Heat supplied by steam if it is to condense totally and convert into water at $100^{\circ} C$ .
$Q_{1} = 1 \times L_{V} = 540 cal$
Heat required by ice to convert totally into water at $0^{\circ} C$
$Q_{1} = 1 \times L_{i c e} = 1 \times 80 = 80 cal$
Heat required by ice to convert totally into water at $100^{\circ} C$
$Q_{3} = 1 \times L_{i c e} + 1 \times C \times Δ T = 1 \times 80 + 1 \times 1 \times (100 - 0) = 80 + 100 = 180 cal$
Since $Q_{1} > Q_{3}$ , entire steam will not condense and the final temperature of mixture of water and steam will be $100^{\circ} C$ .

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Similar questions

Q. One

gram

of ice is mixed with one

gram

of steam. After thermal equilibrium is reached, the temperature of the mixture is
(Latent heat of fusion of ice;

L_{i c e} = 80 cal/g

, latent heat of condensation of vapour;

L_{V} = 540 cal/g

)

600 g

of ice at

253 K

is mixed with

0.08 kg

of steam at

100^{\circ} C

. Latent heat of vaporization of steam

= 540 cal/g

. Latent heat of fusion of ice

= 80 cal/g

. Specific heat of ice

= 0.5

{cal/g/}^{\circ} C

. The resultant temperature of the mixture is

2 g

of steam at

100^{\circ} C

is passed into

6 g

of ice at

0^{\circ} C

. If the latent heat of steam and ice are

540 cal/g

and

80 cal/g

respectively and specific heat capacity of water is

1 cal / g^{\circ} C

, then the final temperature of the mixture is

1 g

of steam at

100^{\circ} C

and an 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

, specific heat of water

= 1 c a l / g^{\circ} C

)

60 g

of ice at

0^{0}

is mixed with

60 g

of steam at

100^{0}

. At thermal equilibrium, the mixture contains:(Latent heats of steam and ice are

540 c a l g^{- 1}

and

80 c a l g^{- 1}

respectively, specific heat of water

= 1 c a l g^{- 1}^{- 1}

)

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Latent Heat

Standard XII Physics

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One gram of ice is mixed with one gram of steam. After thermal equilibrium is reached, the temperature of the mixture is (Latent heat of fusion of ice; Lice=80 cal/g, latent heat of condensation of vapour; LV=540 cal/g, Specific heat capacity of water; c=1 cal/g∘C)