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

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Homework Help Overview

The problem involves calculating the total heat required to change 1 kg of ice at -20 degrees Celsius to steam at 130 degrees Celsius. The discussion centers around the phases of water and the associated heat transfer calculations, including specific heat capacities and latent heats.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants present various calculations for the heat transfer involved in the phase changes of water, questioning the accuracy of their results compared to a teacher's answer. There are discussions about the specific heat capacities and latent heats used in the calculations.

Discussion Status

Some participants are seeking clarification on specific equations and concepts related to latent heat, while others are questioning the definitions of phase changes such as fusion and vaporization. There is an ongoing exploration of the calculations and the underlying principles without a clear consensus on the correct approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific methods or formats for presenting their calculations. There is also a noted discrepancy between their calculations and the expected answer provided by a teacher.

xxiangel
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How much heat must be absorbed by 1 kg of ice at -20 degrees celsius 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 celsius 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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