A 35.41-g sample of Cu(s) at 96.73°C was added to a constant-pressure calorimeter containing 148.3 g of H2O(l) at 25.20°C. The final temperature of the system was 26.66°C. Assuming no heat loss to the surroundings, what is the calorimeter constant? C(Cu) = 24.435 J/mol.K; C(H2O) = 75.291J.K-1.mol-1
To find the calorimeter constant, we need to determine the heat exchanged by the copper (Cu) and water (H2O) during the process.
First, we calculate the heat transferred by the Cu using the formula:
q_cu = m_cu × C_cu × ΔT_cu
Where:
q_cu = heat transferred by copper
m_cu = mass of copper (35.41 g)
C_cu = specific heat capacity of copper (24.435 J/mol.K)
ΔT_cu = change in temperature of copper (final temperature - initial temperature)
ΔT_cu = T_final - T_initial = 26.66°C - 96.73°C
Next, we calculate the heat transferred by the water:
q_h2o = m_h2o × C_h2o × ΔT_h2o
Where:
q_h2o = heat transferred by water
m_h2o = mass of water (148.3 g)
C_h2o = specific heat capacity of water (75.291 J.K-1.mol-1)
ΔT_h2o = change in temperature of water (final temperature - initial temperature)
ΔT_h2o = T_final - T_initial = 26.66°C - 25.20°C
Since the calorimeter is a constant-pressure calorimeter, the heat gained by the water is equal to the heat lost by the copper:
q_h2o = -q_cu
Therefore, we can set the two equations of heat transfer equal to each other:
m_h2o × C_h2o × ΔT_h2o = -m_cu × C_cu × ΔT_cu
Substituting the known values:
148.3 g × 75.291 J.K-1.mol-1 × (26.66 - 25.20)°C = -35.41 g × 24.435 J/mol.K × (26.66 - 96.73)°C
Simplifying the equation:
148.3 g × 75.291 J.K-1.mol-1 × 1.46°C = -35.41 g × 24.435 J/mol.K × -70.07°C
Now, we can solve for the calorimeter constant:
C_calorimeter = q_h2o / ΔT_h2o
Substituting the known values:
C_calorimeter = (148.3 g × 75.291 J.K-1.mol-1 × 1.46°C) / (26.66 - 25.20)°C
Calculating this expression will give us the calorimeter constant.