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.

Explanation:

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 .

tube or goog with be a great resource to with that question

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg