Suppose 3.6 kg of ice at -4˚C is added to an aquarium containing 20 gallons of...
Suppose 3.6 kg of ice at -4˚C is added to an aquarium containing 20 gallons of water initially at 25˚C. What is temperature of the water when all the ice melts and the system is at a constant temperature?
Homework Answers
Mw = mass of water = 20 gallons = 75.71 kg
m = mass of ice = 3.6 kg
L = latent heat of fusion of ice to water = 334000 J/kg
cice = specific heat of ice = 2050
cwater = specific heat of water = 4186
T = final equilibrium temperature
using conservation of energy
m cice (4) + mL + m c (T) = M cwater (25 - T)
(3.6) (2050) (4) + 3.6 x 334000 + (3.6) (4186) (T) = (75.71) (4186) (25 - T)
(3.6) (2050) (4) + 3.6 x 334000 + (3.6) (4186) (T) = (75.71) (4186) (25 - T)
T = 20.2
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