A 73.6 g sample of aluminum is heated to 95.0°C and then dropped into 100.0 g of water initially at 20.0°C. If the water's temperature rises to 30.0°C, what is the specific heat capacity of the metal?

 To determine the specific heat capacity of aluminum, we can 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 will be equal to the heat gained by the water. We can set up the problem using the following formula for heat transfer:

\[ q = mc\Delta T \]

Where: - \( q \) is the heat transferred - \( m \) is the mass of the substance - \( c \) is the specific heat capacity of the substance - \( \Delta T \) is the change in temperature

For the aluminum (which we'll call substance "A"), we have:

\[ q_A = m_Ac_A(T_{initial, A} - T_{final, A}) \]

For the water (substance "W"), we have:

\[ q_W = m_Wc_W(T_{final, W} - T_{initial, W}) \]

Given that the heat lost by the aluminum is equal to the heat gained by the water:

\[ q_A = -q_W \]

Now, we plug in the given values:

- For aluminum, \( m_A = 73.6 \) g, \( T_{initial, A} = 95.0°C \), and \( T_{final, A} = 30.0°C \). - For water, \( m_W = 100.0 \) g, \( c_W = 4.184 \) J/g°C (specific heat capacity of water), \( T_{initial, W} = 20.0°C \), and \( T_{final, W} = 30.0°C \).

Let's calculate \( q_W \):

\[ q_W = m_Wc_W(T_{final, W} - T_{initial, W}) \] \[ q_W = (100.0 \text{g}) * (4.184 \text{J/g°C}) * (30.0°C - 20.0°C) \] \[ q_W = 100.0 * 4.184 * 10 \] \[ q_W = 4184 \text{J} \]

We assume that \( q_A = -q_W \) because the heat lost by the aluminum is the heat gained by the water:

\[ q_A = -4184 \text{J} \]

Now, solve for \( c_A \):

\[ m_Ac_A(T_{initial, A} - T_{final, A}) = -q_W \] \[ 73.6 \text{g} * c_A * (95.0°C - 30.0°C) = -4184 \text{J} \] \[ 73.6 * c_A * 65 = 4184 \] \[ 4784 * c_A = 4184 \] \[ c_A = \frac{4184}{4784} \] \[ c_A = 0.875 \text{J/g°C} \]

Thus, the specific heat capacity of the aluminum is approximately \( 0.875 \text{J/g°C} \).