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Standard XII

Chemistry

Heat of Reaction

The latent he...

Question

The latent heat of vaporisation of water is:

2.25 \times 10^{6} J / k g

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2.25 \times 10^{8} J / k g

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2.25 \times 10^{4} J / k g

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2.25 \times 10^{- 6} J / k g

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Solution

The correct option is A $2.25 \times 10^{6} J / k g$
The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,250 joules (533 calories) per gram or $2.25 * 10^{6}$ J/kg.

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Similar questions

The latent heat of vaporisation of water is:
(a) $2.25 \times 10^{6} J / K g$ (a) $3.34 \times 10^{6} J / K g$
(c) $2.25 \times 10^{4} J / K g$ (d) $33.4 \times 10^{5} J / K g$

The latent heat of fusion of ice is:

(a) 33.4 $\times 10^{5} J / K g$ (b) 22.5 $\times 10^{5} J / K g$
(c) 33.4 $\times 10^{4} J / K g$ (d) 2.25 $\times 10^{4} J / K g$

Q. 1 kg of water is converted into steam at the same temperature and at 1 atm (100 kPa). The density of water and steam are 1000 kgm

^{- 3}

and 0.6 kgm

^{- 3}

respectively. The latent heat of vaporisation of water is

2.25 \times 10^{6} J k g^{- 1}

. What will be increase in energy?

Q. If 1 kg of vegetables costs

$

4.23, then 2.25 kg of vegetables costs .

Q. The increase in internal energy of

1

kg of water at

100^{0} C

when it is converted into steam at the same temperature and at

1 a t m

(100 k P a)

will be [The density of water and steam are

1000 k g / m^{3}

and

0.6 k g / m^{3}

respectively. The latent heat of vapourisation of water is

2.25 \times 10^{6} J / k g

] :

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The latent heat of vaporisation of water is:

The correct option is A 2.25×106J/kgThe heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,250 joules (533 calories) per gram or 2.25∗106 J/kg.

The correct option is A $2.25 \times 10^{6} J / k g$
The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,250 joules (533 calories) per gram or $2.25 * 10^{6}$ J/kg.