The increase in the internal energy of 10 g of water when it is heated from 0- to 100- and converted into steam at 100 kPa is given as x - 102 J- Find x-The density of steam - 0-6 kg-m3- Specific heat capacity of water - 4200 J-kg- C and the latent heat of vaporization of water - 2-5 - 106 J-kg-

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Byju's Answer

Standard XII

Physics

Thermodynamic Processes

The increase ...

Question

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

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Solution

So we can write $Δ U = Δ Q - Δ W$
Heat is supplied in two processes.
$Δ Q_{1}$ to raise the temperature of water from $0^{\circ}$ to $100^{\circ}$
$Δ Q_{2}$ to turn the water at $100^{\circ}$ into steam.
$Δ Q_{1} = m . s . Δ T = 0.01 \times 4200 \times 100 = 4200 J$
$Δ Q_{2} = m L = 0.01 \times 2.5 \times 10^{6} = 25000 J$
$Δ Q = Δ Q_{1} + Δ Q_{2} = 29200 J$
$Δ W = Δ (P V) = P Δ V$ (As expansion occurs at constant pressure of $100 k P a$ )
Initially, the volume of water will be negligible.
After expansion into steam, its volume becomes $\frac{M a s s}{D e n s i t y} = \frac{0.01}{0.6} \approx 0.017 m^{3}$
$Δ W = 100 \times 10^{3} \times 0.017 = 1700 J$
$∴ Δ U = 29200 - 1700 = 27500 J$
Given $27500 = x \times 10^{2}$
Therefore $x = 275$
( $Δ (P V)$ during heating of water can be neglected as it will be very less )

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