When a 290 g piece of iron at 180°C is placed in 1.5 kg of water at 20°C, the final equilibrium temperature can be calculated using principles of heat transfer. This involves equating the heat lost by the iron to the heat gained by the water, considering their specific heat capacities and masses.
Understanding Heat Transfer: Iron and Water Equilibrium
The core concept here is thermal equilibrium. When two objects at different temperatures come into contact, heat naturally flows from the hotter object to the colder one. This process continues until both objects reach the same final temperature. To determine this final temperature, we use the principle of conservation of energy, specifically focusing on heat exchange.
The Physics Behind the Calculation
The amount of heat transferred (Q) is calculated using the formula:
$Q = mc\Delta T$
Where:
- m is the mass of the substance.
- c is the specific heat capacity of the substance.
- $\Delta T$ is the change in temperature ($\Delta T = T_{final} – T_{initial}$).
We will assume that no heat is lost to the surroundings. This means the heat lost by the iron is equal to the heat gained by the water.
Key Values Needed
Before we can calculate, we need some crucial values:
- Mass of iron ($m_{iron}$): 290 g = 0.290 kg
- Initial temperature of iron ($T_{i, iron}$): 180°C
- Mass of water ($m_{water}$): 1.5 kg
- Initial temperature of water ($T_{i, water}$): 20°C
- Specific heat capacity of iron ($c_{iron}$): Approximately 450 J/(kg·°C)
- Specific heat capacity of water ($c_{water}$): Approximately 4186 J/(kg·°C)
Setting Up the Heat Exchange Equation
The heat lost by the iron is:
$Q_{lost, iron} = m_{iron} \cdot c_{iron} \cdot (T_{i, iron} – T_{final})$
The heat gained by the water is:
$Q_{gained, water} = m_{water} \cdot c_{water} \cdot (T_{final} – T_{i, water})$
Since $Q_{lost, iron} = Q_{gained, water}$:
$m_{iron} \cdot c_{iron} \cdot (T_{i, iron} – T_{final}) = m_{water} \cdot c_{water} \cdot (T_{final} – T_{i, water})$
Solving for the Final Temperature ($T_{final}$)
Let’s plug in the values and solve for $T_{final}$:
$0.290 \text{ kg} \cdot 450 \text{ J/(kg·°C)} \cdot (180°C – T_{final}) = 1.5 \text{ kg} \cdot 4186 \text{ J/(kg·°C)} \cdot (T_{final} – 20°C)$
$130.5 \cdot (180 – T_{final}) = 6279 \cdot (T_{final} – 20)$
$23490 – 130.5 T_{final} = 6279 T_{final} – 125580$
Now, let’s rearrange the equation to isolate $T_{final}$:
$23490 + 125580 = 6279 T_{final} + 130.5 T_{final}$
$149070 = 6409.5 T_{final}$
$T_{final} = \frac{149070}{6409.5}$
$T_{final} \approx 23.26°C$
So, when the 290 g piece of iron at 180°C is placed in 1.5 kg of water at 20°C, the final equilibrium temperature will be approximately 23.26°C.
Factors Affecting the Final Temperature
While our calculation provides a good estimate, several real-world factors can influence the actual final temperature:
- Heat Loss to Surroundings: In reality, some heat will always escape into the environment. This would slightly lower the final equilibrium temperature.
- Container Material: The material of the container holding the water also absorbs some heat. We’ve assumed an ideal scenario where only the iron and water exchange heat.
- Phase Changes: If the initial temperatures were much higher or lower, phase changes (like water boiling or iron melting) would significantly alter the heat exchange dynamics and require different calculations.
- Mixing Efficiency: Ensuring thorough mixing of the water is crucial for uniform temperature distribution.
Comparing Specific Heat Capacities
The significant difference in specific heat capacities between iron and water plays a key role. Water has a much higher specific heat capacity, meaning it requires more energy to raise its temperature by one degree compared to iron. This is why the large mass of water only warms up slightly, while the smaller mass of iron cools down considerably.
| Substance | Specific Heat Capacity (J/(kg·°C)) |
|---|---|
| Water | 4186 |
| Iron | 450 |
This table clearly illustrates why the water’s temperature change is less dramatic than the iron’s.
Frequently Asked Questions (PAA)
What is specific heat capacity?
Specific heat capacity is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius (or one Kelvin). It’s a fundamental property that tells us how well a material stores thermal energy. For instance, water’s high specific heat capacity is why it’s used in cooling systems.
How does mass affect heat transfer?
Mass is directly proportional to the amount of heat transferred. A larger mass of a substance requires more energy to change its temperature compared to a smaller mass of the same substance. In our scenario, the much larger mass of water compared to the iron means the water’s temperature changes less drastically.
Why is the final temperature closer to the initial water temperature?
The final temperature is closer to the initial water temperature because water has a significantly higher specific heat capacity than iron. This means water can absorb a large amount of heat energy with only a small increase in its temperature, while iron loses heat rapidly. The large mass of water also contributes to this effect.
Can iron rust when heated?
Iron can oxidize (rust) at high temperatures, especially in the presence of oxygen and moisture. However, the temperatures involved in this
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