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Standard VII

Physics

Conduction

One gram of s...

Question

One gram of steam at $100^{0} C$ is added to m gram of ice at $0^{0} C$ .What is the mass of ice that will melt and reach the temperature of steam water at $100^{0} C$ .
$[L_{i c e} = 80 c a l / g r a m a n d L_{s t e a m} = 540 c a l / g r a m]$

1 g

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2 g

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3 g

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4 g

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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

)

Q. Heat required to convert 1 g of ice at

0^{0} C

into steam at

100^{0} C

is (

L_{S t e a m} = 536 c a l / g

) :

Q. In an energy recycling process,

X g

of steam at

100^{0} C

becomes water at

100^{0} C

which converts

Y g

of Ice at

0^{0} C

into water at

100^{0} C

. The ratio of

X / Y

will be

Q. One gram of ice at

0^{0} C

is added to

5

grams of water at

10^{0} C

. If the latent heat of ice be

80

cal/g, then the final temperature of the mixture is :

Q. Ice at

0^{\circ} C

is added to

1 g

of steam at

100^{\circ} C

. For

x g

of ice added, the temperature of steam reduces to

0^{\circ} C

. Find the value of

x

.
(Latent heat of ice

= 80 cal/g

and latent heat of steam

= 540 cal/g

. Also, specific heat of water

= 1 cal/ kg K

)

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One gram of steam at 1000C is added to m gram of ice at 00C.What is the mass of ice that will melt and reach the temperature of steam water at 1000C.[Lice=80cal/gramandLsteam=540cal/gram]

One gram of steam at $100^{0} C$ is added to m gram of ice at $0^{0} C$ .What is the mass of ice that will melt and reach the temperature of steam water at $100^{0} C$ .
$[L_{i c e} = 80 c a l / g r a m a n d L_{s t e a m} = 540 c a l / g r a m]$