A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The aluminum sample starts off at 90.6°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 32.3°C. The pressure remains constant at 1atm.Calculate the mass of the aluminum sample. Be sure your answer is rounded to the correct number of significant digits.
Question
A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The aluminum sample starts off at 90.6°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 32.3°C. The pressure remains constant at 1atm.Calculate the mass of the aluminum sample. Be sure your answer is rounded to the correct number of significant digits.
Solution
To solve this problem, we need to use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the heat lost by the aluminum is equal to the heat gained by the water.
The formula for heat transfer is Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, we calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water Q_water = 100.0g * 4.18J/g°C * (32.3°C - 25.0°C) Q_water = 3054.2J
Next, we calculate the mass of the aluminum using the heat lost by the aluminum, which is equal to the heat gained by the water:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum 3054.2J = m_aluminum * 0.897J/g°C * (90.6°C - 32.3°C) 3054.2J = m_aluminum * 0.897J/g°C * 58.3°C
Solving for m_aluminum, we get:
m_aluminum = 3054.2J / (0.897J/g°C * 58.3°C) m_aluminum = 57.9g
So, the mass of the aluminum sample is approximately 57.9g.
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