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Based on the thermodynamic properties provided for water, the energy change when the temperature of 0.650 kg of water decreased from 101 °c to 51.0 °c is 1,609.66 kJ
What are thermodynamic properties of water?
Thermal characteristics of water include its density, freezing point, boiling point, latent heat of evaporation, melting point, and critical temperature, among others.
Here we calculate the boiling point of water is 100°C. So at 101°C, the water is steam. The specific heat first from 101 to 100 calculated as
E = mCΔT, where c for steam is 1.996 kJ/kg·°C
E₁ = (0.65 kg)(1.996 kJ/kg·°C)(101 - 100°C) = 1.2974 kJ
The latent heat when steam turns to liquid. The heat of vaporization of water is 2260 kJ/kg.
E₂ = mHvap = (0.65 kg)(2260 kJ/kg) = 1469 kJ
Thus solving the energy to bring down the temperature to 51°C. The specific heat of liquid water is 4.187 kJ/kg·°C.
E₃ = (0.65 kg)(4.187 kJ/kg·°C)(100 - 51°C) = 139.36 kJ
Total energy = 1.2974 kJ+1469 kJ+139.36 kJ = 1,609.66 kJ
Therefore , the energy change when the temperature of 0.650 kg of water decreased from 101 °c to 51.0 °c is 1,609.66 kJ
To know more about thermodynamic properties from the given link
https://brainly.com/question/24969033
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