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Standard XII

Physics

Calorimetry

Calculate the...

Question

Calculate 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.

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Solution

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 $\times 10^{6} J / Kg$
$∆ t = Change in temperature of the system = 100 ° C = 373 K$

∆Q = Heat absorbed to raise the temperature of water from 0 $° C to 100 ° C$ + Latent heat for conversion of water to steam
∆Q = $m c ∆ t + m L$
= 0.01 × 4200 × 100 + 0.01 × 2.5 × 10⁶
= 4200 + 25000 = 29200 J

∆W = P∆V
∆V $= mass (\frac{1}{final density} - \frac{1}{initial density})$
$∆ V = (\frac{0.01}{0.6}) - (\frac{0.01}{1000}) = 0.01699$

∆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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Calculate 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.