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Standard VII

Chemistry

Latent Heat of Vaporisation

How much stea...

Question

How much steam at $100^{o}$ C will melt $64$ gm of ice at $- 10^{o}$ C....(specific heat of ice $= 0.5$ and latent of ice $- 80$ cal/gm)?

8.5

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4.5

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6.5

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2.5

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Solution

The correct option is A $8.5$ gm
As we know that the amount of heat absorbed by the ice is equals to the amount of heat released by the steam.
$m L_{S} + m s_{ω} (100 - 0) = m_{i c e} \times S_{i c e} [0 - (- 10)] + m_{i c e} \times 80$
$m \times 540 + m \times 1 \times 100 = 64 \times 0.5 \times 10 + 64 \times 80$
$640 m = 64 \times 85$
$m = 8.5 g$

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Similar questions

Q. 1 g of a steam at

100^{_{-}^{0}} C

melt how much ice at

0^{_{-}^{0}} C

? (Latent heat of ice = 80 cal/gm and latent heat of steam = 540 cal/gm)-

20 gm

ice at

- 10^{\circ} C

is mixed with

m gm

steam at

100^{\circ} C

. The minimum value of

m

so that finally all ice and steam converts into water at

0^{\circ} C

(

Use: specific heat and latent heat as

C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{f u s i o n}

= 80 cal/gm

and

L_{v a p o r}

= 540 cal/gm

)

20 gm

ice at

- 10^{\circ} C

is mixed with

m gm

steam at

100^{\circ} C

. The minimum value of

m

so that finally all ice and steam converts into water is

[

Use, specific heat and latent heat as

C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{m e l t}

= 80 cal/gm

and

L_{v a p o r}

= 540 cal/gm

]

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

50 g m

ice at -

10^{0} C

is mixed with

20 g m

steam at

100^{0} C .

When the mixture finally reaches its steady state inside a calorimeter of water equivalent

1.5 g m

then : [Assume calorimeter was initially at

0^{0} C,

Take latent heat of vaporization of water

= 540

cal/gm, Latent heat of fusion of water

= 80

cal/gm, specific heat capacity of water

= 1 c a l / g m -^{0} C,

specific heat capacity of ice

= {0.5 cal/ gm}^{0} C

]

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How much steam at 100oC will melt 64 gm of ice at −10oC....(specific heat of ice =0.5 and latent of ice −80 cal/gm)?

The correct option is A 8.5gmAs we know that the amount of heat absorbed by the ice is equals to the amount of heat released by the steam.mLS+msω(100−0)=mice×Sice[0−(−10)]+mice×80m×540+m×1×100=64×0.5×10+64×80640m=64×85m=8.5g

How much steam at $100^{o}$ C will melt $64$ gm of ice at $- 10^{o}$ C....(specific heat of ice $= 0.5$ and latent of ice $- 80$ cal/gm)?