A 45g piece of ice 0.0∘C added to water at 7.4∘C. All ice melts and the temp of the water decreases to 0.0∘C. How many g of water in sample?
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A 45 g piece of ice at 0.0 ∘C is added to a sample of water at 7.4 ∘C. All of the ice melts and the temperature of the water decreases to 0.0 ∘C. How many grams of water were in the sample?
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Energy to melt the ice = 0.045 kg x 334 kJ/kg = 15.03 kJ
that energy cools the water
E = 15.03 = 4.186 kJ/kgC x 7.4º x M
M = 15.03/ 4.186•7.4 = 0.485 kg = 485 g
specific heat of water is 4.186 kJ/kgC = 4.186 J/gC
specific heat of ice is 2.06 kJ/kgC
specific heat of steam is 2.1 kJ/kgK
heat of fusion of ice is 334 kJ/kg
heat of vaporization of water is 2256 kJ/kg
Calculate heat absorbed when ice melts. To do this, you need the heat of fusion of ice which is 334 J/g (Check this value with your text).
q = m (heat of fusion)
q = 45 g (334 J/g) = 1.50X10^4 J
So, the water must have lost 1.50X10^4 J. Using the initial and final temps of the water, and its specific heat, you can calculate the mass of water:
q = m (c) (delta T)
1.50X10^4 J = m (4.184 J/gC) (7.4C)
m = 485 g water