To calculate the heat removal required to convert 2 kg of water at 20°C to ice at -10°C, three stages must be considered: cooling the water from 20°C to 0°C, phase change from water to ice at 0°C, and cooling the ice from 0°C to -10°C. The specific heat of water is 1.0 cal/g°C, and the latent heat of fusion is 80 cal/g. The total heat removal can be calculated using the equation Q=cmT for each stage, leading to a definitive total heat removal value in kcal.
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And the specific heat of water is 1.0 cal/g °C.Yanbeast said:Homework Statement
How much heat (in kcal) must be removed to make ice at –10°C from 2 kg of water at 20°C? (The specific heat of ice is 0.5 cal/g °C.)
Yanbeast said:Homework Equations
Q=cmT
The Attempt at a Solution
I thought this problem was simple,
Q=0.5x2000x30
Then diving by 1000 again to put it into Kcal. 30 is not the right answer.