Calculate the increase in internal energy of 1 kg of water at 100 degrees C when it is converted into - Physics - Thermodynamics

Question

Calculate the increase in internal energy of 1 kg of water at
100 degrees C when it is converted into steam at the same
temperature and at 1atm Te density of water and steam are
1000kg/m^2 and 0 6 kg m^ -3 respectively The Latent heat of
vapourisation of water is 2 25 X 10^6 J kg^ -1

Vijay Yadav · Accepted Answer

Hello,

Given Latent heat of vaporization L_v = 2.25 × 10⁶ J/kg

Mass of the water m = 1kg,

P = 1×10⁵ Pa

Volume of 1 kg water (V)

$V = \frac{1}{1000} m^{3} [∵ m = ρ v]$

Volume of 1kg stream V'

$V' = \frac{1}{0.6} m^{3}$

The increase in volume ΔV = (V' – V)

$= (\frac{1}{0.6} - \frac{1}{1000}) = (1.7 - 0.00) m^{3} \approx 1.7 m^{3}$

The work done by system

ΔW = PΔV

⇒ ΔW = (1 × 10⁵) (1.7)

⇒ ΔW = 1.7 × 10⁵ J

The heat given to convert 1 kg water at 100°c into 1 kg steam at 100°c

ΔQ = mL_V

= 1 × 2.25 × 10⁶ = 2.25 × 10⁶ J

The change in internal energy is

ΔU = ΔQ – ΔW

= 2.25 × 10⁶– 1.7 × 10⁵

= (2.25 – 0.17) × 10⁶

ΔU = 2.08 × 10⁶ J

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Calculate the increase in internal energy of 1 kg of water at 100 degrees C when it is converted into steam at the same temperature and at 1atm.Te density of water and steam are 1000kg/m^2 and 0.6 kg m^ -3 respectively. The Latent heat of vapourisation of water is 2.25 X 10^6 J kg^ -1.