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) $33.4 \times 10^{5} J / K g$

Question

The latent heat of vaporisation of water is-2-25times 10-6 J-kg2-25times 10-8 J-kg2-25times 10-4 J-kg2-25times 10-6- J-kg

Toppr Admin · Accepted Answer

The heat of fusion for water at 0 -xB0-C is approximately 334 joules -79-7 calories- per gram- and the heat of vaporization at 100 -xB0-C is about 2-250 joules -533 calories- per gram or 2-25-x2217-106 J-kg-

The latent heat of vaporisation of water is:
$2.25 \times 10^{6} J / k g$
$2.25 \times 10^{8} J / k g$
$2.25 \times 10^{4} J / k g$
$2.25 \times 10^{- 6} J / k g$

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.

The latent heat of vaporisation of water is:2.25×106J/kg2.25×108J/kg2.25×104J/kg2.25×10−6J/kg

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∗106 J/kg.

The latent heat of vaporisation of water is:
$2.25 \times 10^{6} J / k g$
$2.25 \times 10^{8} J / k g$
$2.25 \times 10^{4} J / k g$
$2.25 \times 10^{- 6} J / k g$

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.