Calculate the increase in entropy of 5 kg of water at 100^0C when it changes to vapour
Answers
Answer:
This question doesn’t specify the water’s final state (100 C vapor or liquid?) and assigns an incorrect value to liquid water’s specific heat ( “heat capacity”) - it’s 4180 J per kg, not mole
The liquid’s entropy change would be its heat capacity (Cp) times the natural log of the two absolute temperatures or 4.18 J/g degree K*18 g/mol *(ln*((273+100)/(273+90)= 2.0447 J/(K*mol)
For 10 kg (10000/18 moles) of water, that ΔS works out to 1136 J/K
Since it takes about 540 calories (or 4.18*540 J) to evaporate 1 gram of water at atmospheric pressure and all of that energy goes into increasing its entropy, that change would add another 60,500 (10,000*(4.18*540/373) J/K
Consequently if the water ends up as steam. its total entropy change would 1136+60500 or 61700 J/degree.
Explanation:
Answer: The change in entropy is 61700 j/degree.
Explanation: The final state of the water at 100 C is not stated in this question (is it a liquid or vapour?). and gives the inaccurate estimate of 4180 J per kilogramme (not mole) for the specific heat of liquid water.
The heat capacity (Cp) times the natural logarithm of the two absolute temperatures, or 4.18 J/g degree K*18 g/mol *(ln*((273+100)/(273+90)= 2.0447 J/(K*mol)), would represent the liquid's entropy change.
S equals 1136 J/K for 10 kg (10000/18 moles) of water.
Since 1 gramme of water evaporates at atmospheric pressure using 540 calories (4.18*540 J), all of that energy is used to increase the entropy of the water, which results in an additional 60,500 (10,000*(4.18*540/373) J/K) of change.
As a result, if the water turns into steam. the change in entropy would be 61700 j/degree.