Using these equations with appropriate values for specific heat of ice, water, a
ID: 581397 • Letter: U
Question
Using these equations with appropriate values for specific heat of ice, water, and steam, and enthalpies the of fusion and vaporization, we have: q total = (m-c-AT) ice + n-AHfus + (m-c-AT)water + n-AHvap + (m-c-AT)steam = (135 g.209 J/g°C-15°C) + (135-Th 02L-6.01 kJ/mol 18.02 g +(135 g-4.18 J/g-C-1000) + (135 g.1802L-40.67 kJ/mol +(135 g.1.84 J/g.°C-20°C) = 4230 J + 45.0 kJ + 56,500 J + 305 kJ + 4970 Converting the quantities in J to kJ perm its them to be summed, yielding the total heat required: =4.23 kJ +45.0 kJ + 56.5 kJ + 305 kJ + 4.97 kJ = 416 kJ Check Your Learning What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at -30.0 °c? Answer: 40.5 kJ 10.4 Phase Diagrams By the end of this section, you will be able to: about us careers privacy policy terms of use contact us helpExplanation / Answer
Here you have to add three values of heat:
The sensible heat that the water releases with formula m*Cp*Change in temperature, m is mass and Cp is the specific heat 4.184 g/C for water, this is the heat release by liquid water until it reaches 0C
the latent hea which is mass * latent heat fusion of water wich is 6.01 KJ / mol, this is the heat the water needs to release to change its physical form, no change of temperature here
the sensible heat that water releases to go from 0 to -30 in its solid phase
so
heat of liquid water from 80 to 0 C
q = 94 g * 4.18 J /gC * (0 - 80) = -31,433.6 Joules, the change of temperature is (Final temp - Initial temp)
the negative sign is because the water is losing heat
heat to change physical form
latent heat is 6.01 KJ / mol, we need to have KJ / g, divide thisvalue by 18.01
6.01 / 18 = 0.3338 KJ / g , multiply by 1000 to get J, so latent heat is 333.8 J / gr
and finallly the sensible heat of ice going from 0 to -30 with a specific heat of 2.09 J/ g C
latent heat for our particular case
q = 333.8 * 94 = 31,377.2 Joules
sensible heat for ice going from 0 to - 30
q = m * Specific heat * change in temperature
q = 94 * 2.09 * (-30 - 0) = -5893.8 Joules
negative sign is because the lost of energy
total heat is
q = 31,433.6 Joules + 31,377.2 Joules + 5, 893.8 Joules
q released = 68704.6 Joules
q released = 68.704 KJ
I know it constrasts the answer from your textbook but this is how you properly solve this questions
hope it helps !!
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