Section 7.4: Heat Capacity and Mixture
When two substances at different temperatures are mixed, heat is transferred from the hotter to the cooler substance until thermal equilibrium is reached. This is the principle of calorimetry.
Using specific heat capacities, we can calculate the final equilibrium temperature (\( T_f \)):
Where:
\( m_1, m_2 \) = masses of the substances (kg)
\( c_1, c_2 \) = specific heat capacities (J/kg·K)
\( T_1, T_2 \) = initial temperatures (°C or K)
\( T_f \) = final equilibrium temperature
Example 1: Mixing Water
100 g of water at 80 °C is mixed with 200 g of water at 20 °C. Find the final temperature. Assume \( c = 4180 \, \text{J/kg·K} \).
\[
m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)
\]
\[
0.1 \cdot 4180 (80 - T_f) = 0.2 \cdot 4180 (T_f - 20)
\]
\[
8.36 - 0.418 T_f = 0.836 T_f - 16.72
\]
\[
25.08 = 1.254 T_f \Rightarrow T_f \approx 20 \, °C
\]
The final temperature is approximately 46 °C.
Example 2: Metal in Water
A 0.2 kg piece of metal at 100 °C is placed into 0.5 kg of water at 25 °C. The metal has \( c = 400 \, \text{J/kg·K} \) and water \( c = 4180 \, \text{J/kg·K} \). Find the final temperature.
\[
m_\text{metal} c_\text{metal} (T_\text{metal} - T_f) = m_\text{water} c_\text{water} (T_f - T_\text{water})
\]
\[
0.2 \cdot 400 (100 - T_f) = 0.5 \cdot 4180 (T_f - 25)
\]
\[
80 (100 - T_f) = 2090 (T_f - 25)
\]
Solve for \( T_f \approx 27.8 \, °C \)
Practice Problems
- 50 g of water at 90 °C is mixed with 150 g of water at 30 °C. Find the final temperature.
- 0.3 kg of metal at 150 °C is dropped into 0.4 kg of water at 25 °C. Metal's \( c = 380 \, \text{J/kg·K} \). Find \( T_f \).
- A 200 g aluminum block at 80 °C is placed into 100 g of water at 20 °C. Calculate \( T_f \).
- Why is the mass of the substances important in determining the final temperature?
- Two liquids, 0.5 kg and 0.3 kg, are mixed. Their specific heat capacities are 4000 J/kg·K and 2000 J/kg·K. Initial temperatures: 80 °C and 20 °C. Find \( T_f \).
- An unknown metal is dropped into water, and the final temperature is measured. How can you determine its specific heat capacity?
- Explain the assumptions made in calorimetry problems.
- Calculate the heat lost by the hotter substance in problem 1.
- A mixture reaches equilibrium at 50 °C. One substance had mass 0.4 kg and specific heat 1000 J/kg·K. The other had mass 0.2 kg. Find its specific heat capacity if initial temperatures were 70 °C and 20 °C.
- Why is calorimetry important in chemistry and physics experiments?