The amount of heat required to convert 1 g of ice specific 0-5 cal at g-1o C-1 at -100 C to steam at 100 - C is - Given- Latent heat of ice is 80 Cal- gm- Latent heat of steam is 540 Cal-gm- Specific heat of water is 1 Cal-gm-C -

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Standard X

Physics

Power

The amount of...

Question

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 $1 C a l / g m / C$ ]

725 cal

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636 cal

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716 cal

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None of these

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Solution

The correct option is A 725 cal
Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water at 100 C to vapor at 100 C
$= 1 \times 0.5 [0 - (- 10)] + 1 \times 80 + 1 \times 1 \times 100 + 1 \times 540 = 5 + 80 + 100 + 540 = 725 c a l$

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Similar questions

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

)

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

]

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

)

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 at

- 10^{0} C

into steam at

100^{0} C

is (approximately): (specific heat of water

=

4.187

J/g

^{0}

C, latent heat of fusion

=

334

J/g , latent heat of vaporization

=

2260

J/g, specific heat of ice

=

2.03

J/g) :

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The amount of heat required to convert 1 g of ice (specific 0.5 cal at g−1oC−1 ) at −100C to steam at 100 ∘C is ___________.[ Given: Latent heat of ice is 80Cal/gm, Latent heat of steam is 540Cal/gm, Specific heat of water is 1Cal/gm/C ]

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 $1 C a l / g m / C$ ]