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

Byju's Answer

Standard XII

Physics

Calorimetry

1 kg of water...

Question

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?

3 \times 10^{5} J

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

4 \times 10^{6} J

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

2.08 \times 10^{6} J

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

None of these

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

Open in App

Solution

The correct option is C $2.08 \times 10^{6} J$
Volume of $1 k g$ of water $= \frac{m}{ρ}$ $= \frac{1 k g}{1000 k g / m^{3}}$ $= 0.001 m^{3}$

Volume of $1 k g$ of steam $= \frac{m}{ρ^{'}}$ $= \frac{1 k g}{0.6 k g / m^{3}}$ $= \frac{10}{6} m^{3}$

Hence, work done by system $= P Δ V$
$= 100 k P a \times 1.73 m^{3}$
$= 1.7 \times 10^{5} J$
Heat given $= m L = 2.25 \times 10^{6} J$
Hence, increase in energy of water $= 2.25 \times 10^{6} J - 1.7 \times 10^{5} J$
$= 2.08 \times 10^{6} J$

Suggest Corrections

Similar questions

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 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. The increase in internal energy of 1 kg of water at

100^{0} C

when it is converted into steam at the same temperture and at 1 atm (100 kPa) 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

10^{6}

J/kg.)

Q. The increase in internal energy of 1 kg of water at

100^{0} C

when it is converted into steam at the same temperture and at 1 atm (100 kPa) 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

10^{6}

J/kg.)

Calculate the increase in the internal energy of 10 g of water when it is heated from $0^{\circ} C$ to $100^{\circ} C$ and converted into steam at 100 kPa. The density of steam $= 0.6 k g m^{- 3}$ . Specific heat capacity of water $= 4200 J k g^{- 1}^{\circ} C^{- 1}$ and the latent heat of vaporization of water $= 2.25 \times 106 J k g^{- 1}$ .

Join BYJU'S Learning Program

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 2.25×106Jkg−1. What will be increase in energy?

The correct option is C 2.08×106JVolume of 1 kg of water =mρ =1kg1000kg/m3 =0.001m3Volume of 1 kg of steam =mρ′ =1kg0.6kg/m3 =106m3Hence, work done by system =PΔV =100kPa×1.73m3 =1.7×105JHeat given =mL=2.25×106JHence, increase in energy of water =2.25×106J−1.7×105J =2.08×106J

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?