A cube of ice is taken from the freezer at - 5.5 ∘C and placed in a 75-g aluminum calorimeter filled with 300 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? Express your answer to two significant figures and include the appropriate units.

Question

Luna Phillips

Physics

31 May, 12:24

A cube of ice is taken from the freezer at - 5.5 ∘C and placed in a 75-g aluminum calorimeter filled with 300 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? Express your answer to two significant figures and include the appropriate units.

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Answers (1)

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Elodie · Answer 1

Let mass of ice cube taken out be m kg.

ice will gain heat to raise its temperature from - 5.5° to 0° and then from 0° to 17°.

Total heat gained = m x 2.1 x 5.5 + m x 333 + m x 4.186 x 17

= (11.55 + 333 + 71.162) m

= 415.712 m kJ

Heat lost by aluminium calorimeter

=.075 x. 9 x 3

=.2025 kJ

Heat lost by water

=.3 x 4.186 x 3

= 3.7674

Total heat lost

= 3.9699 kJ

Heat lost = heat gained

415.712 m = 3.9699

m =.0095 kg

9.5 gm.