1g of water, of volume 1 cm$^3$ at 100?C, is converted into steam at same temperature under normal atmospheric pressure ($\simeq$ ) $1\times 10^5$Pa . The volume of steam formed equals 1671 cm$^3$. If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is,

Question

Explain why :

a body with large reflectivity is a poor emitter
a brass tumbler feels much colder than a wooden tray on a chilly day
an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
the earth without its atmosphere would be inhospitably cold
heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Answer 1

To find the change in internal energy when 1 gram of water is converted into steam, we can use the first law of thermodynamics, which is given by the formula:

Q = \Delta U + W

where:

Q is the heat added to the system
\Delta U is the change in internal energy
W is the work done by the system

Given:

Specific latent heat of vaporization of water, L = 2256 \text{ J/g}
Heat added (Q) = 2256 \text{ J} (since 1 gram of water is converted to steam)
Change in volume, \Delta V = 1671 \text{ cm}^3 - 1 \text{ cm}^3 = 1670 \text{ cm}^3
Pressure, P = 1 \times 10^5 \text{ Pa}

Convert \Delta V to \text{m}^3:

\Delta V = 1670 \times 10^{-6} \text{ m}^3 = 1.67 \times 10^{-3} \text{ m}^3

The work done by the system during expansion is given by:

W = P \Delta V

Substitute the values to find W:

W = 1 \times 10^5 \times 1.67 \times 10^{-3} = 167 \text{ J}

Now, substitute Q and W in the first law of thermodynamics to find \Delta U:

\Delta U = Q - W = 2256 - 167 = 2089 \text{ J}

Therefore, the change in internal energy is 2089 J. Thus, the correct option is 2089 J.

Answer 2