Sure, let's solve the problem step by step:
### Step 1: Define the Given Data
- Mass of the aluminum sample, [tex]\( m_{\text{Al}} \)[/tex]: [tex]\( 0.400 \)[/tex] kg
- Specific heat capacity of aluminum, [tex]\( c_{\text{Al}} \)[/tex]: [tex]\( 9.10 \times 10^2 \)[/tex] J/(kg⋅°C)
- Initial temperature of aluminum, [tex]\( T_{\text{initial, Al}} \)[/tex]: [tex]\( 95.0 \)[/tex] °C
- Mass of the water, [tex]\( m_{\text{water}} \)[/tex]: [tex]\( 0.550 \)[/tex] kg
- Specific heat capacity of water, [tex]\( c_{\text{water}} \)[/tex]: [tex]\( 4184 \)[/tex] J/(kg⋅°C)
- Initial temperature of water, [tex]\( T_{\text{initial, water}} \)[/tex]: [tex]\( 18.0 \)[/tex] °C
### Step 2: Set Up the Heat Transfer Equation
When the aluminum is dropped into the water, heat is transferred from the aluminum to the water until thermal equilibrium is reached. At equilibrium, the mixture will come to a final temperature [tex]\( T \)[/tex].
The heat lost by the aluminum will be equal to the heat gained by the water:
[tex]\[ Q_{\text{lost by Al}} = Q_{\text{gained by Water}} \][/tex]
Using the formula for heat transfer [tex]\( Q = mc\Delta T \)[/tex], where [tex]\( \Delta T = T_{\text{initial}} - T \)[/tex] for the substance losing heat and [tex]\( \Delta T = T - T_{\text{initial}} \)[/tex] for the substance gaining heat, we have:
[tex]\[ m_{\text{Al}} \cdot c_{\text{Al}} \cdot (T_{\text{initial, Al}} - T) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]
### Step 3: Substitute the Known Values
[tex]\[ 0.400 \cdot 9.10 \times 10^2 \cdot (95.0 - T) = 0.550 \cdot 4184 \cdot (T - 18.0) \][/tex]
### Step 4: Simplify and Solve for [tex]\( T \)[/tex]
Let's simplify the equation:
[tex]\[ 0.400 \cdot 910 \times (95.0 - T) = 0.550 \cdot 4184 \cdot (T - 18.0) \][/tex]
[tex]\[ 364 \cdot (95.0 - T) = 2301.2 \cdot (T - 18.0) \][/tex]
Expanding both sides:
[tex]\[ 364 \cdot 95.0 - 364T = 2301.2T - 2301.2 \cdot 18.0 \][/tex]
[tex]\[
364 \cdot 95.0 = 34580, \quad 2301.2 \cdot 18.0 = 41421.6
\][/tex]
[tex]\[
34580 - 364T = 2301.2T - 41421.6
\][/tex]
### Step 5: Combine Like Terms and Solve for [tex]\( T \)[/tex]
Combining the like terms:
[tex]\[
34580 + 41421.6 = 2301.2T + 364T
\][/tex]
[tex]\[
76001.6 = 2665.2T
\][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[
T = \frac{76001.6}{2665.2}
\][/tex]
[tex]\[
T \approx 28.52 \text{ °C}
\][/tex]
Therefore, the final temperature of the mixture will be approximately [tex]\( 28.52 \)[/tex] °C.
