Physics, 23.07.2019 15:40 noname5359

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.650 kg of water decreased from 101 °c to 51.0 °c.

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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

Answer from: evandlubbep6bsvu

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, $T_{i}=101^{\circ}\rm C$ .

The final temperature of water is, $T_{f}=51^{\circ}\rm C$ .

Since, the boiling point of water is $100 ^{\circ} \rm C$ , 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,

$\Delta E= mc(T_{f}-T_{i}) +mh$

Here, c is the specific heat of water and h is the latent heat of vaporization of water. Its value is $2257 \;\rm kJ/kg$ .

Solving as,

$\Delta E= 0.65 \times 4.187 \times (51-101) +(0.65 \times 2257)\\\Delta E = 1333.69 \;\rm kJ$

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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t. w e n t y c h a r a c t e r s

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