A(n) 80-g ice cube at 0°C is placed in 750 g of water at 22°C. What is the final temperature of the mixture?
_________C
Heat of fusion of ice = 334 J/g
Specific heat of water = 4.186 J/g*˚C
heat energy is required to melt all the ice.
The water’s temperature will decrease as the ice melts.
Energy absorbed by ice to melt = mass * Heat of fusion = 80 * 334 =
26,720 J
2nd determine the temperature of the water, after releasing the
heat energy.
Energy released by water = Mass * Specific heat * ∆T
Energy released by water = 750 * 4.18 * ∆T
750 * 4.186 * ∆T = 80 * 334
∆T = 3.19˚
The water’s temperature decreased 3.19˚ as all the ice melted.
Final temperature of water = 22 - 3.19 = 18.81
Now you have 750 grams of water at 18.81˚C and 80 grams of water at
0˚
3rd determine the final temperature of the water.
80 * 4.186 * (Tf – 0) = 750 * 4.186 * (18.81– Tf)
Divide both sides by 4.186
80 * Tf = 750 * (18.81 – Tf)
830 * Tf = 14107.5
Tf = 16.99˚
final temp is 16.99
A(n) 80-g ice cube at 0°C is placed in 750 g of water at 22°C. What...
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