OUNT OF HEAT NEEDED TO CHANGE A SUBSTANCE FROM ONE PHASE TO ANOTHER

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In summary, the amount of heat required for 1 kg of ice at -20 degrees Celcius to change to steam at 130 degrees is 3,100,000 J. This is calculated by adding the heat needed for ice to reach 0 degrees (41200 J), the heat needed to melt ice (334,000 J), the heat needed for water to reach 0 degrees (334,000 J), the heat needed to vaporize water (2,260,000 J), and the heat needed for steam to reach 130 degrees (262,600 J). It is important to note that the specific latent heats for fusion and vaporization are different and must be used accordingly in the calculations.
  • #1
xxiangel
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How much heat must be absorbed by 1 kg of ice at -20 degrees Celcius to change it to steam at 130 degrees?

This is the work that I came up with.

Ice -20°C to Ice 0°C= q=mc(change in T)= 1kg x 2060 J/kg.Kx 20 degrees= 41200 J

Ice 0 degrees to Water 0 degrees= Q= Hf x m= 334,000 j/kg x 1 kg= 334,000 J

Water 0 degrees to steam 0 degrees= Q= Hf x m= 334,000 j/kg x 1 kg= 334,000 J

Steam 0 degrees to steam 130 degrees= q=mc(change T) = 1 kg x 2020 j/kg x 130 degrees= 262,600

The final answer I get is 971,800 Joules. The teacher says the answers is 3,100,000 J.

What the heck am I doing wrong?? can you help me?
:confused: :confused: :confused: :confused: :confused: :confused: :cry:
 
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  • #2
xxiangel said:
How much heat must be absorbed by 1 kg of ice at -20 degrees Celcius to change it to steam at 130 degrees?

This is the work that I came up with.

Ice -20°C to Ice 0°C= q=mc(change in T)= 1kg x 2060 J/kg.Kx 20 degrees= 41200 J

Ice 0 degrees to Water 0 degrees= Q= Hf x m= 334,000 j/kg x 1 kg= 334,000 J

Water 0 degrees to steam 0 degrees= Q= Hf x m= 334,000 j/kg x 1 kg= 334,000 J

Steam 0 degrees to steam 130 degrees= q=mc(change T) = 1 kg x 2020 j/kg x 130 degrees= 262,600

The final answer I get is 971,800 Joules. The teacher says the answers is 3,100,000 J.
There are 5 different heats here:

[tex]Q_1 = mc_i\Delta T_1[/tex] (20 deg. change, c_i = 2060 J/K-Kg.)

[tex]Q_2 = mL_f[/tex] (L_f = 334,000 J/Kg.)

[tex]Q_3 = mc_w\Delta T_2[/tex] (100 deg. change, c_w = 4184 J/K-Kg.)

[tex]Q_4 = mL_v[/tex] (L_v = 2,260,000 J/Kg.)

[tex]Q_5 = mc_s\Delta T_3[/tex] (30 deg. change, c_s = 2020 J/K-Kg.)

AM
 
  • #3
xxiangel said:
What does [tex]Q_2 = mL_f[/tex]
mean? Where does this equation come from?
L_f is the specific latent heat of fusion of ice: 334,000 J/Kg.

L_v is the specific latent heat of vaporization of water.

AM
 
  • #4
One more question Andrew, you have been so hopeful already.

How is water to steam considered vaporization in this problem? I thought water to steam was considered fusion, and steam to water was vaporization. Thats how the example problems are in my physics book. This has to do with Q4.
 
  • #5
xxiangel said:
One more question Andrew, you have been so hopeful already.

How is water to steam considered vaporization in this problem? I thought water to steam was considered fusion, and steam to water was vaporization. Thats how the example problems are in my physics book. This has to do with Q4.
Water to steam is vaporization. Fusion is water-ice. Read your text again. Think about it. The energy is required to break the hydrogen bonds between water molecules (vaporization) is the same amount of energy that is given up when those bonds form (when steam condenses).

AM
 

FAQ: OUNT OF HEAT NEEDED TO CHANGE A SUBSTANCE FROM ONE PHASE TO ANOTHER

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Temperature problem refers to any issue or challenge related to measuring, regulating, or understanding temperature in a specific system or environment. It may involve determining the cause of temperature fluctuations, finding ways to control temperature, or optimizing temperature for a certain purpose.

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