Latent heat of vaporisation of water is 22.6 × 105 J / kg. The amount of heat needed to convert 100 kg of water into vapor at 100 ° C is

Question

Latent heat of vaporisation of water is 22.6 × 105 J / kg. The amount of heat needed to convert 100 kg of water into vapor at 100 ° C is (A) 11.3 × 105 J (B) 11.3 × 106 J (C) 22.6 × 106 J (D) 22.6 × 107 J

Tardigrade Admin · Accepted Answer

Heat required to convert water at 100° C into steam at 100° C is Q=m L where, m= mass =100 kg and L= latent heat =226 × 105 J / kg ∴ Q=100 × 22.6 × 105=22.6 × 107 J

Q. Latent heat of vaporisation of water is $22.6 \times 1 0^{5} J / k g$ . The amount of heat needed to convert $100 k g$ of water into vapor at $100^{\circ} C$ is

$11.3 \times 1 0^{5} J$

$11.3 \times 1 0^{6} J$

$22.6 \times 1 0^{6} J$

$22.6 \times 1 0^{7} J$

Heat required to convert water at $10 0^{\circ} C$ into steam at $10 0^{\circ} C$ is $Q = m L$
where, $m =$ mass $= 100 k g$ and
$L =$ latent heat $= 226 \times 1 0^{5} J / k g$
$∴ Q = 100 \times 22.6 \times 1 0^{5} = 22.6 \times 1 0^{7} J$

Q. Latent heat of vaporisation of water is 22.6×105J/kg. The amount of heat needed to convert 100kg of water into vapor at 100∘C is

11.3×105J

11.3×106J

22.6×106J

22.6×107J