Heat required in converting 1 g of ice at -10- C into steam at 100- C is-latent heat of fusion- 80 cal-g80 cal-glatent heat of vaporization-540 cal-g

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Byju's Answer

Standard XII

Physics

1st Law of Thermodynamics

Heat required...

Question

Heat required in converting $1 g$ of ice at $- 10^{\circ} C$ into steam at $100^{\circ} C$ is:
latent heat of fusion= 80 cal/g
latent heat of vaporization=540 cal/g

3045 J

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6056 J

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721 J

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6 J

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Solution

The correct option is A $3045 J$
We know specific heat of ice is $0.5 c a l / g K$ , specific heat of water=1cal/gK
Also latent heat of fusion= $80 c a l / g$
latent heat of vaporization= $540 c a l / g$
The following processes occur during the above said conversion.
$I c e (263 K) \to i c e (273 K) \to w a t e r (273 K) \to w a t e r (373 K) \to s t e a m (373 K)$
Thus total heat absorbed= $(0.5 \times 10 + 80 + 1 \times 100 + 540) c a l = 725 c a l = 3045 J$

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Similar questions

Q. The amount of heat required to convert

1 g

of ice at

- 10^{o}

C into steam at

100^{o}

C is (specific heat of ice

= 0.5

cal/g/

^{o}

C, latent heat of ice

= 80

cal/g, latent heat of steam

= 540

cal/g)

Q. Work done in converting one gram of ice at

- 10^{\circ} C

into steam at

100^{\circ} C

is -
[Given, specific heat of ice

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

, specific heat of water

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

, latent heat of fusion of ice

L_{f} = 80 cal/g

, & latent heat of vaporization of steam

L_{v} = 540 cal/g

]

Q. The heat required (in

c a l

) to change 10 g ice at

0^{o} C

to steam at

100^{o} C

is:
[ Heat of fusion and heat of vaporization for

H_{2} O

are

80 c a l / g

and

540 c a l / g

respectively. Specific heat of water is

1 c a l / g

]

Ratio of heat energy required to change in states of $100 g$ of water at $100^{\circ} C$ to steam to $540 g$ of ice to water respectively is $5 : y$ then $y$ = ___

Take latent heat of fusion $80 c a l / g m$ and latent heat of vaporization as $540 c a l / g m$

Property	Value
Heat of Fusion (cal/g)	80
Heat of Vaporization (cal/g)	540
Specific Heat (cal/g. $^{o}$ C)	1.00

For conversion of 10 grams of ice at 0

^{o}

C to liquid water at 100

^{o}

C heat needed is:

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Heat required in converting 1g of ice at −10∘C into steam at 100∘C is:latent heat of fusion= 80 cal/g80 cal/glatent heat of vaporization=540 cal/g

Heat required in converting $1 g$ of ice at $- 10^{\circ} C$ into steam at $100^{\circ} C$ is:
latent heat of fusion= 80 cal/g
latent heat of vaporization=540 cal/g