NEET-XII-Physics
26: Laws of Thermodynamics
- #20Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m-3. Specific heat capacity of water = 4200 J kg-1 °C-1 and the latent heat of vaporization of water = 2.25 × 10 6J kg-1.Ans : Given:
Mass of water, m = 10 g = 0.01 kg
Pressure, P = 105 Pa
Specific heat capacity of water, c = 1000 J/Kg `` °C``
Latent heat, L = 2.25`` \times {10}^{6}\,\mathrm{\,J\,}/\,\mathrm{\,Kg\,}``
`` ∆t=\,\mathrm{\,Change\,}\,\mathrm{\,in\,}\,\mathrm{\,temperature\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,system\,}=100°\,\mathrm{\,C\,}=373\,\mathrm{\,K\,}``
∆Q = Heat absorbed to raise the temperature of water from 0`` °\,\mathrm{\,C\,}\,\mathrm{\,to\,}100°\,\mathrm{\,C\,}`` + Latent heat for conversion of water to steam
∆Q = `` mc∆t+mL``
= 0.01 × 4200 × 100 + 0.01 × 2.5 × 106
= 4200 + 25000 = 29200 J
∆W = P∆V
∆V`` =\,\mathrm{\,mass\,}\left(\frac{1}{\,\mathrm{\,final\,}\,\mathrm{\,density\,}}-\frac{1}{\,\mathrm{\,initial\,}\,\mathrm{\,density\,}}\right)``
`` ∆\,\mathrm{\,V\,}=\left(\frac{0.01}{0.6}\right)-\left(\frac{0.01}{1000}\right)=0.01699``
∆W = P∆V = 0.01699 × 105 = 1699 J
Using the first law, we get
∆Q = ∆W + ∆U
∆U = ∆Q - ∆W = 29200 - 1699
= 27501 = 2.75 × 104 J
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