1. Apr 23, 2005

### xxiangel

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?

2. Apr 23, 2005

### Andrew Mason

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

3. Apr 23, 2005

### Andrew Mason

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. Apr 23, 2005

### xxiangel

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. Apr 23, 2005

### Andrew Mason

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