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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.

User Suvonkar
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2 Answers

2 votes

Final answer:

The mass of the ice cube can be found by equating the heat gained by the ice while warming, melting, and heating to the final temperature with the heat lost by the water and aluminum calorimeter cooling to the final temperature.

Step-by-step explanation:

To determine the mass of the ice cube, we must first understand that energy will be exchanged between the ice, the aluminum calorimeter, and the water until thermal equilibrium is reached. The ice must first warm up to 0 °C, melt, and then raise in temperature to the final temperature of 17.0 °C. The calorimeter and water will cool down to this same temperature. The heat gained by the ice will equal the heat lost by the water and calorimeter combined.

Let's denote the mass of the ice as m, in kilograms. The amount of heat required to warm the ice from -5.5 °C to 0 °C is given by q1 = m × c_ice × ΔT, where c_ice is the specific heat of ice. To melt the ice at 0 °C, we use q2 = m × L_fusion, where L_fusion is the heat of fusion of ice. Finally, to warm the melted ice from 0 °C to 17.0 °C, we need q3 = m × c_water × ΔT, where c_water is the specific heat of water.

The heat lost by the water is q4 = m_water × c_water × ΔT, and the heat lost by the calorimeter is q5 = m_cal × c_al × ΔT. The sum of the heat lost by water and the calorimeter must equal the sum of the heats gained by the ice in all its transformations, so q1 + q2 + q3 = q4 + q5. By substituting the given values and solving for m, we get the mass of the ice.

User Beygi
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4.8k points
0 votes

Answer:

0.00954g or 9.5x
10^(-3) kg

Step-by-step explanation:

The only conversion that is needed is changing the heat of fusion of water from 333 kJ/kg to 333000 J/kg.

This is the condensed version of the equation needed for this problem: mcΔT + mL + mcΔT = mcΔT + mcΔT

This is the expanded version of the equation needed for this problem:


m_(ice)(
c_(ice))(temperature of ice from -5.5°C to 0°C) +
m_(ice)(L) +
m_(ice)(
c_(water))(temperature of water from 0°C to 17°C) =
m_(water)(
c_(water))(ΔT) +
m_(aluminum)(
c_(aluminum))(ΔT)

Use the equation to solve for the mass of ice:

m(2100)(5.5) + m(333000) + m(4186)(17) = 0.3(4186)(20-17) + 0.075(900)(20-17)

m [(2100x5.5) + 333000 + (4186x17)] = 3767.4 + 202.5

m(415712) = 3969.9

m = 0.00954g or 9.5x
10^(-3) kg

User Mingwei Zhang
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