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Effect of Change of Temperature on State Change of Matter

Latent heat o...

Question

Latent heat of fusion for ice is :-

A

80 gm cal^–1

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B

80 cal / gm

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C

19 J cal^–1

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D

20 J cal^–1

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Solution

The correct option is B
80 cal / gm

Latent heat of fusion for ice is 80 cal / gm.

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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)-

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

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

- 10^{0} C

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

20 gm

- 10^{\circ} C

m gm

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

L_{v a p o r}

= 540 cal/gm

)

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

50 g m

10^{0} C

20 g m

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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