Section 7.5: Heat Capacity and Calorimetry
Calorimetry is the study of heat transfer between substances to determine quantities such as heat capacity and specific heat. It relies on the principle that heat lost by a hot object equals heat gained by a cooler object:
The heat absorbed or released by a substance is given by:
Where:
\( Q \) = heat (J)
\( m \) = mass (kg)
\( c \) = specific heat capacity (J/kg·K)
\( \Delta T \) = temperature change (K or °C)
Example 1: Mixing Water and Metal
A 0.3 kg piece of copper at 100 °C is placed in 0.5 kg of water at 25 °C. Specific heat capacities: copper \( 385 \, \text{J/kg·K} \), water \( 4180 \, \text{J/kg·K} \). Find the final temperature \( T_f \).
\[
m_\text{Cu} c_\text{Cu} (T_\text{Cu} - T_f) = m_\text{water} c_\text{water} (T_f - T_\text{water})
\]
\[
0.3 \cdot 385 (100 - T_f) = 0.5 \cdot 4180 (T_f - 25)
\]
Solve for \( T_f \approx 31.6 \, °C \)
Example 2: Determining Specific Heat
A 0.2 kg unknown metal at 90 °C is dropped into 0.3 kg water at 20 °C. The final temperature is 26 °C. Determine the specific heat of the metal.
\[
m_\text{metal} c_\text{metal} (T_\text{initial} - T_f) = m_\text{water} c_\text{water} (T_f - T_\text{water})
\]
\[
0.2 \, c_\text{metal} (90 - 26) = 0.3 \cdot 4180 (26 - 20)
\]
\[
0.2 \, c_\text{metal} \cdot 64 = 0.3 \cdot 4180 \cdot 6
\]
\[
c_\text{metal} = \frac{0.3 \cdot 4180 \cdot 6}{0.2 \cdot 64} \approx 586.875 \, \text{J/kg·K}
\]
Practice Problems
- A 150 g aluminum block at 80 °C is placed in 200 g water at 25 °C. Find the final temperature. \( c_\text{Al} = 900 \, \text{J/kg·K} \).
- Determine the specific heat of a metal if 0.25 kg at 100 °C is dropped into 0.5 kg of water at 20 °C and the final temperature is 30 °C.
- 100 g of water at 60 °C is mixed with 200 g at 30 °C. Calculate the final temperature.
- A mixture reaches equilibrium at 40 °C. One substance had mass 0.4 kg and \( c = 800 \, \text{J/kg·K} \), the other 0.2 kg. Find its specific heat if its initial temperature was 90 °C.
- Explain why calorimetry experiments assume no heat loss to surroundings.
- How does the mass ratio affect the final temperature in mixtures?
- 0.3 kg of metal at 150 °C is placed in 0.4 kg water at 25 °C. The final temperature is 50 °C. Find the specific heat of the metal.
- A liquid at 80 °C is mixed with an equal mass of water at 20 °C. Final temperature is 50 °C. Find the specific heat of the liquid.
- Explain the significance of thermal equilibrium in calorimetry.
- A 0.2 kg metal piece at 120 °C is placed in 0.3 kg water at 25 °C. If \( T_f = 35 °C \), find the heat lost by the metal and gained by water.