A piece of aluminum was placed in a calorimeter containing water. The temperature of the water increased, indicating that the aluminum transferred heat to the water and the calorimeter. The specific heat of the calorimeter was calculated to be 0.874 J/g°C.
When a piece of aluminum weighing 52 grams, and at a
temperature of 76.3 oC, is placed in a calorimeter containing 75.0
grams of water at 21.9oC, the temperature of the water is increases to
28.3oC.
Step 1: Calculate the heat transfer of water and aluminum
The heat transfer of water (Q_water) can be calculated using the following equation:
Q_water = m_water * c_water * ΔT_water
where:
m_water is the mass of water (75.0 g)
c_water is the specific heat of water (4.184 J/g°C)
ΔT_water is the change in temperature of water (28.3°C - 21.9°C = 6.4°C)
Therefore, the heat transfer of water is:
Q_water = 75.0 g * 4.184 J/g°C * 6.4°C = 1942.24 J
The heat transfer of aluminum (Q_aluminum) can be calculated using the following equation:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
where:
m_aluminum is the mass of aluminum (52.0 g)
c_aluminum is the specific heat of aluminum (0.902 J/g°C)
ΔT_aluminum is the change in temperature of aluminum (21.9°C - 76.3°C = -54.4°C)
Therefore, the heat transfer of aluminum is:
Q_aluminum = 52.0 g * 0.902 J/g°C * -54.4°C = -2351.36 J
Step 2: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter (Q_calorimeter) can be calculated using the following equation:
Q_calorimeter = -(Q_water + Q_aluminum)
where:
Q_water is the heat transfer of water (1942.24 J)
Q_aluminum is the heat transfer of aluminum (-2351.36 J)
Therefore, the heat absorbed by the calorimeter is:
Q_calorimeter = -(1942.24 J - 2351.36 J) = 409.12 J
Step 3: Calculate the specific heat of the calorimeter
The specific heat of the calorimeter (c_calorimeter) can be calculated using the following equation:
c_calorimeter = Q_calorimeter / (m_water * ΔT_water)
where:
Q_calorimeter is the heat absorbed by the calorimeter (409.12 J)
m_water is the mass of water (75.0 g)
ΔT_water is the change in temperature of water (6.4°C)
Therefore, the specific heat of the calorimeter is:
c_calorimeter = 409.12 J / (75.0 g * 6.4°C) = 0.874 J/g°C