10 g of ice at 0- C is mixed with 5 g of steam at 1000- C- If latent heat of fusion of ice is 80 cal - g and latent heat of vaporization of 540 cal - g- Then at thermal equilibriumA- Temperature of the mixture is 0- CB- Temperature of mixture is 100- CC- Mixture contains 13-3 g of water and 1-67 g of steamD- Mixture contains 5-3 g of ice and 9-7 g of water

The correct options are B Temperature of mixture is 100^∘C C Mixture contains 13.3g of water and 1.67 g of steam Heat gain by ice to melt =mL_f = 10 × 80 = 800 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Specific Heat Capacity

10 g of ice a...

Question

10 g of ice at $0^{\circ} C$ is mixed with 5g of steam at $1000^{\circ} C$ . If latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of 540 cal/g. Then at thermal equilibrium

Temperature of the mixture is

0^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Temperature of mixture is

100^{\circ} C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Mixture contains 13.3g of water and 1.67 g of steam

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Mixture contains 5.3g of ice and 9.7g of water

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
B Temperature of mixture is $100^{\circ} C$
C Mixture contains 13.3g of water and 1.67 g of steam
Heat gain by ice to melt = $m L_{f} = 10 \times 80 = 800$ cal
Heat gain by 10 g water to raise its temperature from $0^{\circ} C$ to $100^{\circ} C = 10 \times 1 \times 100 = 1000 c a l$
Mass of steam converted into water
$∴ 1800 = m \times 540 \Rightarrow m = 3.33 g$
$\Rightarrow$ Equilibrium temperature $100^{\circ} C$
Amount of water 10 + 3.33 = 13.33 g
Amount of steam = 5 – 3.33 = 1.67 g

Suggest Corrections

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

)

5 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

)

1 kg

of ice at

- 10^{\circ} C

is mixed with

4.4 kg

of water at

30^{\circ} C

. The final temperature of mixture is

Given specific heat of ice

= 0.5 {cal/g}^{\circ} C

,
Specific heat of water

= 1 {cal/g}^{\circ} C

,
Latent heat of fusion for ice

= 80 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

Join BYJU'S Learning Program

10 g of ice at 0∘C is mixed with 5g of steam at 1000∘C. If latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of 540 cal/g. Then at thermal equilibrium

10 g of ice at $0^{\circ} C$ is mixed with 5g of steam at $1000^{\circ} C$ . If latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of 540 cal/g. Then at thermal equilibrium