Which contains more heat, 1 kg of water at 100° C or 1 kg of steam at 100° C ? Give reason for your answer.

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Solution

Latent heat of vaporisation of water is 22.5 × 10⁵ J/kg. It means that 22.5 × 10⁵ J of heat is required to change 1 kg of water at 100°C into steam at the same temperature. It means that 1 kg of steam at 100°C has 22.5 × 10⁵ J more energy than 1 kg of water at the same temperature.

In other words, when the temperature of water is raised, it starts getting converted into steam, as soon as the latent heat of vaporization is achieved. Since, water gets converted into steam on gaining heat, steam has more heat than water at the same temperature, in this case, 100°C. Steam gets converted into water on loosing this heat.

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Similar questions

Q. Which contains more heat, 1 kg of ice at 0° C or 1 kg of water at 0° C? Give reason for your answer.

1 kg

ice at

- 20^{\circ} C

is mixed with

1 kg

steam at

200^{\circ} C

. The equilibrium temperature and mixture content is

In following equation calculate the value of H.

1kg steam at $20 0^{0} C = H + 1 K g$ water at $100^{0} C$

$(S_{s t e a m} = C o n tan t = .5 c a l / g m^{0} C)$

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$

1 k g

of ice at

0^{\circ} C

is mixed with

1 k g

of steam at

100^{\circ} C

. What will be the composition of the system when thermal equilibrium is reached? Latent of fusion

3.36 \times 10 J / g

and latent heat of vaporization of water =

2.26 \times 10 J / g

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