You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Latent Heat of Vaporization

Find the heat...

Question

Find the heat needed to convert 10 g of water into vapour (Latent heat of vapourisation of water = $2.24 \times 10^{6} J {k g}^{- 1}$ )

2.20 \times 10^{6} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.30 \times 10^{56} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.24 \times 10^{6} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.24 \times 10^{4} J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2.24 \times 10^{4} J$
Given , $L_{w} = 2.24 \times 10^{6} J / k g, m = 10 g = 0.01 k g$
From the definition of latent heat ,
$Q = m L$
or $Q = 0.01 \times 2.24 \times 10^{6} = 2.24 \times 10^{4} J$

Suggest Corrections

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

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

] :

Q. The increase in the internal energy of

10 g

of water when it is heated from

0^{\circ} t o 100^{\circ}

and converted into steam at

100 k P a

is given as

x \times 10^{2} J

. Find

x

.
The density of steam

= 0.6 k g / m^{3} .

Specific heat capacity of water

= 4200 J / k g -^{\circ} C

and the latent heat of vaporization of water

= 2.5 \times 10^{6} J / k g .

Q. Calculate the increase in internal energy of

1

k g

of water at

100^{\circ} C

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

1

a t m

(

100

k P a

). 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

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

Q. Calculate the increase in the internal energy of

20 g

of water when it is heated from

0^{\circ} C

100^{\circ} C

and converted into steam at

100 kPa

. The density of steam

= 0.50 {kg/m}^{3}

, specific heat capacity of water

= 4200 {J kg}^{- 1}^{\circ} C^{- 1}

and latent heat of vaporization of water

= 2.3 \times 10^{6} J/kg

Join BYJU'S Learning Program

Find the heat needed to convert 10 g of water into vapour (Latent heat of vapourisation of water = 2.24×106Jkg−1)

The correct option is D 2.24×104JGiven , Lw=2.24×106J/kg,m=10g=0.01kgFrom the definition of latent heat , Q=mLor Q=0.01×2.24×106=2.24×104J

Find the heat needed to convert 10 g of water into vapour (Latent heat of vapourisation of water = $2.24 \times 10^{6} J {k g}^{- 1}$ )

The correct option is D $2.24 \times 10^{4} J$
Given , $L_{w} = 2.24 \times 10^{6} J / k g, m = 10 g = 0.01 k g$
From the definition of latent heat ,
$Q = m L$
or $Q = 0.01 \times 2.24 \times 10^{6} = 2.24 \times 10^{4} J$