● A 156.0 g sample of a metal at 75.0°C is placed in 144.6 g of water at 20.0 °C. The temperature of the final system (metal and water) is 23.3 C. What is the specific heat of the metal? specific heat of water = 4.184 J/g C
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![### Problem Statement:
A 156.0 g sample of a metal at 75.0°C is placed in 144.6 g of water at 20.0°C. The temperature of the final system (metal and water) is 23.3°C.
### Question:
What is the specific heat of the metal?
### Given Data:
- Mass of the metal sample (m₁): 156.0 g
- Initial temperature of the metal (T₁_initial): 75.0°C
- Mass of the water (m₂): 144.6 g
- Initial temperature of the water (T₂_initial): 20.0°C
- Final temperature of the system (T_final): 23.3°C
- Specific heat of water (c_water): 4.184 J/g°C
### Steps to Solve:
1. **Calculate the heat gained or lost:**
The heat gained or lost by a substance can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q \) is the heat energy (in Joules, J)
- \( m \) is the mass (in grams, g)
- \( c \) is the specific heat capacity (in J/g°C)
- \( \Delta T \) is the change in temperature (in °C)
2. **Write the heat balance equation:**
Since the system is isolated, the heat lost by the metal will be equal to the heat gained by the water:
\[ Q_{\text{metal}} = -Q_{\text{water}} \]
3. **Calculate ΔT for metal and water:**
\[ \Delta T_{\text{metal}} = T_{\text{final}} - T_{\text{initial, metal}} \]
\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} \]
4. **Substitute the values:**
For water:
\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]
\[ Q_{\text{water}} = 144.6 \text{ g} \times 4.184 \text{ J/g°C} \times (23.3°C - 20.0°C) \]
For metal:
\[ Q_{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78db42f9-6d62-48c9-a117-b436e469106b%2F7a9e7f9d-535b-4da0-a4ee-21aa7a8c6a5b%2Fv59ag4h_processed.jpeg&w=3840&q=75)
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