Finding Final Temperature: Copper In Water

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Finding Final Temperature: Copper in Water

Hey guys! Ever wondered what happens when you drop a hot piece of copper into cool water? Let's dive into a fun and practical chemistry problem where we'll figure out the final temperature when a 95.0 g sample of copper, initially heated to 82.4°C, is placed into water at 22.0°C. Buckle up, because we're about to get into some heat transfer!

Understanding the Basics of Heat Transfer

Before we jump into the nitty-gritty calculations, it's super important to understand the basic concepts of heat transfer. At its core, heat transfer is the process where thermal energy moves from one place to another. This usually happens because of a temperature difference. In our case, the copper is hot, and the water is cool, so the heat will naturally flow from the copper to the water until they reach the same temperature. This final temperature is what we're trying to find.

Specific Heat Capacity

A key concept here is specific heat capacity, often denoted as c. This tells us how much energy it takes to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin, since the size of a degree is the same). Different materials have different specific heat capacities. For example, water has a relatively high specific heat capacity (around 4.18 J/g°C), which means it takes a lot of energy to heat up water compared to, say, copper, which has a lower specific heat capacity (around 0.20 J/g°C). This is why coastal areas have milder temperatures than inland areas – the ocean's high heat capacity helps to regulate the temperature.

Heat Transfer Equation

The amount of heat transferred (q) can be calculated using the equation:

q = mcΔT

Where:

  • q is the heat transferred (in Joules)
  • m is the mass of the substance (in grams)
  • c is the specific heat capacity of the substance (in J/g°C)
  • ΔT is the change in temperature (in °C), which is the final temperature (Tfinal) minus the initial temperature (Tinitial)

So, ΔT = Tfinal - Tinitial

In our problem, the heat lost by the copper will be equal to the heat gained by the water, assuming no heat is lost to the surroundings. This is based on the principle of conservation of energy.

Setting Up the Problem

Alright, let's get our ducks in a row. We have a 95.0 g sample of copper ($c_{Cu} = 0.20 J/g°C$) that's heated to 82.4°C. We're dropping this into some water ($c_{H2O} = 4.18 J/g°C$) that's sitting at 22.0°C. The big question is: what's the final temperature when everything settles down?

Known Values

First, let's list out all the known values:

For Copper (Cu):

  • Mass ($m_{Cu}$) = 95.0 g
  • Specific heat capacity ($c_{Cu}$) = 0.20 J/g°C
  • Initial temperature ($T_{i,Cu}$) = 82.4°C

For Water (H2O):

  • Specific heat capacity ($c_{H2O}$) = 4.18 J/g°C
  • Initial temperature ($T_{i,H2O}$) = 22.0°C

Unknown Values

And here's what we need to find:

  • Final temperature ($T_f$) = ? (This will be the same for both copper and water once they reach thermal equilibrium.)

We also need to make an assumption about the mass of the water. Since the mass of water isn't explicitly given, let's assume we have 100.0 g of water. This will make our calculations easier to follow. So,

  • Mass of water ($m_{H2O}$) = 100.0 g

The Heat Exchange Equation

Now, here’s the key idea: the heat lost by the copper equals the heat gained by the water. Mathematically, we can write this as:

qCu=−qH2Oq_{Cu} = -q_{H2O}

Why the negative sign? Because the heat lost by the copper is a negative value (since it's losing energy), and we want the heat gained by the water to be a positive value (since it's gaining energy). The negative sign ensures that the magnitudes are equal but with opposite signs.

Using our heat transfer equation ($q = mcΔT$), we can expand this to:

mCu⋅cCu⋅(Tf−Ti,Cu)=−[mH2O⋅cH2O⋅(Tf−Ti,H2O)]m_{Cu} \cdot c_{Cu} \cdot (T_f - T_{i,Cu}) = -[m_{H2O} \cdot c_{H2O} \cdot (T_f - T_{i,H2O})]

Solving for the Final Temperature

Now, let's plug in all the values we know:

(95.0 g)⋅(0.20 J/g°C)⋅(Tf−82.4°C)=−(100.0 g)⋅(4.18 J/g°C)⋅(Tf−22.0°C)(95.0 \text{ g}) \cdot (0.20 \text{ J/g°C}) \cdot (T_f - 82.4°C) = -(100.0 \text{ g}) \cdot (4.18 \text{ J/g°C}) \cdot (T_f - 22.0°C)

Let's simplify this step by step:

19.0⋅(Tf−82.4)=−418⋅(Tf−22.0)19.0 \cdot (T_f - 82.4) = -418 \cdot (T_f - 22.0)

Expand both sides:

19.0Tf−1565.6=−418Tf+919619.0T_f - 1565.6 = -418T_f + 9196

Now, let's get all the $T_f$ terms on one side and the constants on the other:

19.0Tf+418Tf=9196+1565.619.0T_f + 418T_f = 9196 + 1565.6

Combine like terms:

437Tf=10761.6437T_f = 10761.6

Finally, solve for $T_f$:

Tf=10761.6437≈24.62°CT_f = \frac{10761.6}{437} \approx 24.62°C

So, the final temperature of the water and copper is approximately 24.62°C.

Wrapping It Up

There you have it! When we drop that hot copper into the water, the final temperature settles at around 24.62°C. Remember, this is based on our assumption of 100.0 g of water. If the mass of water were different, the final temperature would also be different. Isn't it cool how we can predict these things using basic physics and chemistry principles?

Key Takeaways:

  • Heat always flows from hotter objects to cooler objects.
  • Specific heat capacity is crucial in determining how much the temperature of a substance changes when it gains or loses heat.
  • The heat lost by one object is equal to the heat gained by another object (assuming no heat loss to the surroundings).

Hope this helps you understand heat transfer a bit better. Keep experimenting and exploring the fascinating world of chemistry!