Find the Mass of Ice Needed to Cool 800mL Water to 0.0°C

AI Thread Summary
To cool 800 mL of water from 28°C to 0.0°C using ice at 0.0°C, the mass of ice needed can be calculated using the formula Q = mc(delta)T. The specific heat capacity of water is 4.184 J/g K, and the temperature change (delta T) is -28°C. The mass of the water can be determined by using the density of water, which is 1000 kg/m³, to convert the volume to mass. The total energy required to cool the water must equal the energy absorbed by the melting ice. This approach will yield the mass of ice necessary to achieve the desired temperature.
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Homework Statement


Ice at 0.0°C is mixed with 8 × 102 mL of water at 28°C. How much ice must melt to lower the water temperature to 0.0°C?


Homework Equations



Q = mc(delta)T

The Attempt at a Solution



c = 4.184 J/g K
(delta)T = -28C
V = 800 mL

I'm not sure what to do since the volume is given and not the mass in this problem.
 
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HINT: What is the density of water?
 
ok, but now how do you find how much ice is needed? when density = mass/v and the density of water is 1000kg/m^3 =mass / 800mL
 
You first need to work out how much energy is required to change the temperature of 816 mL of water by 28oC.
 

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