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Standard X

Physics

Principle of Calorimetry

The amount of...

Question

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) :

3034 J

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735 J

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1000 J

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4200 J

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Solution

The correct option is A 3034 J
Heat required $=$ Ice from -10 to 0 $+$ Ice from 0 to water at 0 $+$ water from 0 to 100 $+$ water at 100 to steam at 100
$= 1 [(2.05) (10) + 334 + (4.187) (100 - 0) + 2260] = 3033.2 J$

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Similar questions

Q. Calculate the total amount of heat energy required to convert

100 g

of ice at

- 10^{\circ} C

completely into water at

100^{\circ} C

. Specific heat capacity of ice =

2.1 J g^{- 1} K^{- 1}

, specific heat capacity of water =

4.2 J g^{- 1} K^{- 1}

, specific latent heat of ice =

336 J g^{- 1}

Q. How much heat energy is released when

5 g

of water at

20^{o} C

changes to ice at

0^{o} C

?
[Specific heat capacity of water

= 4.2 J g^{- 1} C^{- 1}

.
Specific latent heat of fusion of ice

= 336 J g^{- 1}

]

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. Calculate the mass of ice required to lower the temperature of

300 g

of water at

40 ° C

to water at

0 ° C

. (Specific latent heat of ice = 336 J /g, Specific heat capacity of water = 4.2 J/ g°C).

Q. Calculate the amount of heat required to convert

100

g of ice at

0^{o} C

into the water at

0^{o} C

: Given:Specific latent heat of fusion of ice =

336 J / g

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The amount of heat required to convert 1 g of ice at −100C into steam at 1000C is (approximately): (specific heat of water = 4.187 J/g0C, latent heat of fusion = 334 J/g , latent heat of vaporization = 2260 J/g, specific heat of ice = 2.03 J/g) :

The correct option is A 3034 JHeat required = Ice from -10 to 0 + Ice from 0 to water at 0 + water from 0 to 100 + water at 100 to steam at 100 =1[(2.05)(10)+334+(4.187)(100−0)+2260]=3033.2J

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) :

The correct option is A 3034 J
Heat required $=$ Ice from -10 to 0 $+$ Ice from 0 to water at 0 $+$ water from 0 to 100 $+$ water at 100 to steam at 100
$= 1 [(2.05) (10) + 334 + (4.187) (100 - 0) + 2260] = 3033.2 J$