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

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To change 1 kg of ice at -20°C to steam at 130°C, the total heat required is calculated through several phases: heating ice from -20°C to 0°C, melting ice to water, heating water to steam, and then heating steam to 130°C. The calculations provided initially resulted in 971,800 Joules, while the correct total is 3,100,000 Joules. Key components include the specific heats of ice, water, and steam, as well as the latent heats of fusion and vaporization. Clarifications were made regarding the terms "fusion" and "vaporization," emphasizing that fusion refers to the ice-water transition, while vaporization pertains to the water-steam transition. Understanding these concepts is crucial for accurate thermal calculations.
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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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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:

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

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

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

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

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

AM
 
xxiangel said:
What does Q_2 = mL_f
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
 
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.
 
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
 
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