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A cube of ice is taken from the freezer at -7.5 ∘C and placed in a 95-g aluminum calorimeter filled with 330 g of water at room temperature of 20.0 ∘C. The final situation is observed to be all water at 17.0 ∘C. The specific heat of ice is 2100 J/kg⋅C∘, the specific heat of aluminum is 900 J/kg⋅C∘, the specific heat of water is is 4186 J/kg⋅C∘, the heat of fusion of water is 333 kJ/Kg.
What was the mass of the ice cube?
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Let the ice cube's mass be m kg.
The energy needed to warm m kg of ice from -7.5 °C to 0 °C is (7.5 C°)(2100 J/kg·C°)(m kg) = 15 750m J.
The energy needed to melt m kg of ice is (333 kJ/kg)(m kg) = 333m kJ = 333 000m J.
The energy needed to warm m kg of water from 0 °C to 17 °C is (17 C°)(4186 J/kg·C°)(m kg) = 71 162m J.
The total energy input to the ice cube is then 15 750m + 333 000m + 71 162 m J, or 419 912m J.
The energy extractable from 330 g of water, per C°, is (4186 J/kg·C°)(0.33 kg)(1 C°) = 1381.38 J, so the energy made available from the 3 C° loss of temperature is 4144.14 J.
A temperature change of -3 C° extracts (900 J/kg·C°)(3 C°)(0.095 kg) = 427.5 J from the calorimeter.
The extracted energy total is 4 571.64 J.
Divide this by the energy absorbed by the ice, giving m = 0.0109 kg. Since only 2 significant figures are justified here, report it as 0.011 kg or 11 g.