Answer :
The boiling point of water is 100°C. So at 101°C, the water is steam. Compute the specific heat first from 101 to 100.
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
Next, let's solve 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
Lastly, let's solve 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
Thus,
Total energy = 1.2974 kJ+1469 kJ+139.36 kJ = 1,609.66 kJ
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
Next, let's solve 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
Lastly, let's solve 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
Thus,
Total energy = 1.2974 kJ+1469 kJ+139.36 kJ = 1,609.66 kJ
The energy change of water during the change in temperature is 1333.69 kJ
Given data:
The mass of water is, m = 0.650 kg.
The initial temperature of water is, [tex]T_{i}=101^{\circ}\rm C[/tex].
The final temperature of water is, [tex]T_{f}=51^{\circ}\rm C[/tex].
Since, the boiling point of water is [tex]100 ^{\circ} \rm C[/tex], which means it has to convert from steam at 101 Degree Celsius to liquid at 51 Degree Celsius. Therefore, the energy change is given as,
[tex]\Delta E= mc(T_{f}-T_{i}) +mh[/tex]
Here, c is the specific heat of water and h is the latent heat of vaporization of water. Its value is [tex]2257 \;\rm kJ/kg[/tex].
Solving as,
[tex]\Delta E= 0.65 \times 4.187 \times (51-101) +(0.65 \times 2257)\\\Delta E = 1333.69 \;\rm kJ[/tex]
Thus, the energy change of water during the change in temperature is 1333.69 kJ.
Learn more about the latent heat here:
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