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Standard XII

Physics

1st Law of Thermodynamics

Calculate the...

Question

Calculate the increase in the internal energy of $20 g$ of water when it is heated from $0^{\circ} C$ to $100^{\circ} C$ and converted into steam at $100 kPa$ . The density of steam $= 0.50 {kg/m}^{3}$ , specific heat capacity of water $= 4200 {J kg}^{- 1}^{\circ} C^{- 1}$ and latent heat of vaporization of water $= 2.3 \times 10^{6} J/kg$

54.4 kJ

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4 kJ

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50 kJ

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60 kJ

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Solution

The correct option is C $50 kJ$
From $1^{s t}$ law of thermodynamics,
$Q = Δ U + W$
or $Δ U = Q - W$
Now, net heat supplied to the system
$Q = m C Δ T + m L$
$= (\frac{20}{1000} \times 4200 \times 100) + \frac{20}{1000} \times 2.3 \times 10^{6}$
$= 8400 + 46000 = 54400 J$
Work done $W = P Δ V = P (V_{2} - V_{1}) = P (\frac{m}{ρ_{2}} - \frac{m}{ρ_{1}})$
$= 100 \times 10^{3} [\frac{0.02}{0.5} - \frac{0.02}{1000}]$
$\approx 4000 J$
[density of water $= 1000 {kg/m}^{3}$ ]
So, $Δ U = Q - W$
$= 54400 - 4000 = 50400$
$= 5.04 \times 10^{4} J$
$\approx 50 kJ$

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Q. The increase in the internal energy of

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1st Law of Thermodynamics

Standard XII Physics

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Calculate the increase in the internal energy of 20 g of water when it is heated from 0∘C to 100∘C and converted into steam at 100 kPa. The density of steam =0.50 kg/m3, specific heat capacity of water =4200 J kg−1∘C−1 and latent heat of vaporization of water =2.3×106 J/kg