Problem 14.34 Part A A cube of ice is taken from the freezer at-65C and placed i

Question

Problem 14.34 Part A A cube of ice is taken from the freezer at-65C and placed in a 85-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 16.0 °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. | mValue Units Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

Let us assume that mass of the ice cube = m

Initial temperature of ice =-6.5 C

final temperature of ice = 16 C

so heat gained by ice = heat gained from ice at -6.5 C to ice at 0 C + heat gained from ice at 0 C to water at 0 C + heat gained from water at 0 C to water at 16 C

= mass of ice x specific heat of ice x (0-(-6.5) + mass x heat of fusion + mass x specific heat of water x (16-0)

= m x 2.1 x 6.5 + m x 333 + m x 4.186 x 16 = m x 413.626

In this process, while heat was gained by ice at the same time heat was lost by aluminium and water already present.

so heat lost by aluminium and water

= heat lost by aluminium + heat lost by water

= mass of aluminium x specific heat of aluminium x temperature difference + mass of water x specific heat of water x temperature difference

= 0.085 x 0.9 x (20-16) + 0.33 x 4.186 x (20-16)

= 5.83152

As heat gained by ice = heat lost by aluminium and water ( no heat transfer with the surroundings)

so, m x 413.626 = 5.81352

so, m=0.014 kg or 14 g

so mass of the ice cube =14g

all the best in the course work

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