- #1
xxiangel
- 5
- 0
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?
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?