Calorimetry: Finding Final Temperature After Adding Ice

Hey guys! Let's dive into an exciting problem involving heat transfer, phase changes, and thermal equilibrium. Ever wondered what happens when you toss some ice into a container of warm water? Today, we're breaking down a classic calorimetry problem step-by-step. So, grab your thinking caps, and let’s get started!

Problem Statement

Here’s the scenario: We have a copper container, nicely insulated, weighing in at 305 grams. Inside this container, there are 1.5 kilograms of water. Both the container and the water are chilling together at a cozy 38°C. Now, we decide to drop in 550 grams of ice that's at a frosty -10°C. The big question is: What will be the final temperature of the whole system once everything settles down and reaches thermal equilibrium?

Breaking Down the Problem

Calorimetry is all about measuring the heat exchanged in a system. In this case, we need to consider several key processes:

1. Heating the Ice to 0°C: The ice starts at -10°C and needs to reach its melting point.
2. Melting the Ice: The ice needs to transform from a solid to liquid at 0°C.
3. Heating the Melted Ice (Water): The melted ice (now water) will heat up from 0°C to the final temperature.
4. Cooling the Water: The initial 1.5 kg of water will cool down from 38°C to the final temperature.
5. Cooling the Copper Container: The copper container will also cool down from 38°C to the final temperature.

To solve this, we'll use the principle of conservation of energy, which states that the total heat lost by the warmer objects equals the total heat gained by the cooler objects. In simpler terms, the heat lost by the water and copper equals the heat gained by the ice.

Step-by-Step Solution

1. Define the Variables

Let's start by listing all the known values and converting them to consistent units. This will help keep our calculations clear and organized. It’s like setting up your workspace before starting a big project!

• Mass of copper container (${m_c}$): 305 grams = 0.305 kg
• Mass of water (${m_w}$): 1.5 kg
• Initial temperature of water and copper (${T_i}$): 38°C
• Mass of ice (${m_i}$): 550 grams = 0.550 kg
• Initial temperature of ice (${T_{ice}}$): -10°C
• Specific heat of copper (${c_c}$): 385 J/kg°C
• Specific heat of water (${c_w}$): 4186 J/kg°C
• Specific heat of ice (${c_{ice}}$): 2100 J/kg°C
• Latent heat of fusion of ice (${L_f}$): 3.34 × 10⁵ J/kg
• Final temperature (${T_f}$): ? (This is what we want to find!)

2. Calculate the Heat Required to Warm the Ice to 0°C

The first step for the ice is to warm up to its melting point. We use the formula:

${Q = m \, c \, \Delta T}$

Where:

• (${Q}$ is the heat energy,
• (${m}$ is the mass,
• (${c}$ is the specific heat,
• (${ \Delta T}$ is the change in temperature.

So, for the ice:

${Q_1 = m_{ice} \, c_{ice} \, (0 - (-10))}$

${Q_1 = 0.550 \, \text{kg} \, \times 2100 \, \text{J/kg°C} \, \times 10 \, \text{°C}}$

${Q_1 = 11550 \, \text{J}}$

3. Calculate the Heat Required to Melt the Ice at 0°C

Now that the ice is at its melting point, it needs to change from solid to liquid. We use the formula:

${Q = m \, L_f}$

Where:

• (${Q}$ is the heat energy,
• (${m}$ is the mass,
• (${L_f}$ is the latent heat of fusion.

So, for melting the ice:

${Q_2 = m_{ice} \, L_f}$

${Q_2 = 0.550 \, \text{kg} \, \times 3.34 \, \times 10^5 \, \text{J/kg}}$

${Q_2 = 183700 \, \text{J}}$

4. Calculate the Heat Required to Warm the Melted Ice (Water) to the Final Temperature

Once the ice has melted, the resulting water needs to warm up from 0°C to the final temperature (${T_f}$). We use the same formula as before:

${Q = m \, c \, \Delta T}$

So, for warming the melted ice:

${Q_3 = m_{ice} \, c_w \, (T_f - 0)}$

${Q_3 = 0.550 \, \text{kg} \, \times 4186 \, \text{J/kg°C} \, \times T_f}$

${Q_3 = 2302.3 \, T_f \, \text{J}}$

5. Calculate the Heat Lost by the Water as it Cools to the Final Temperature

The initial 1.5 kg of water will cool down from 38°C to the final temperature (${T_f}$). Again, we use the formula:

${Q = m \, c \, \Delta T}$

So, for the water cooling:

${Q_4 = m_w \, c_w \, (38 - T_f)}$

${Q_4 = 1.5 \, \text{kg} \, \times 4186 \, \text{J/kg°C} \, \times (38 - T_f)}$

${Q_4 = 6279 \, (38 - T_f) \, \text{J}}$

${Q_4 = 238602 - 6279 \, T_f \, \text{J}}$

6. Calculate the Heat Lost by the Copper Container as it Cools to the Final Temperature

The copper container also cools down from 38°C to the final temperature (${T_f}$). Using the same formula:

${Q = m \, c \, \Delta T}$

So, for the copper cooling:

${Q_5 = m_c \, c_c \, (38 - T_f)}$

${Q_5 = 0.305 \, \text{kg} \, \times 385 \, \text{J/kg°C} \, \times (38 - T_f)}$

${Q_5 = 117.425 \, (38 - T_f) \, \text{J}}$

${Q_5 = 4462.15 - 117.425 \, T_f \, \text{J}}$

7. Apply the Conservation of Energy Principle

According to the conservation of energy, the total heat gained by the ice equals the total heat lost by the water and copper. Therefore:

${Q_1 + Q_2 + Q_3 = Q_4 + Q_5}$

Substitute the values we calculated:

${11550 + 183700 + 2302.3 \, T_f = 238602 - 6279 \, T_f + 4462.15 - 117.425 \, T_f}$

8. Solve for the Final Temperature (${T_f}$)

Combine the constants and the terms with (${T_f}$):

${195250 + 2302.3 \, T_f = 243064.15 - 6396.425 \, T_f}$

Move the terms with (${T_f}$) to one side and the constants to the other:

${2302.3 \, T_f + 6396.425 \, T_f = 243064.15 - 195250}$

${8698.725 \, T_f = 47814.15}$

Now, divide to find (${T_f}$):

${T_f = \frac{47814.15}{8698.725}}$

${T_f ≈ 5.49 \, \text{°C}}$

Conclusion

So, after all that calculating, we find that the final temperature of the system, once the ice has melted and everything has reached thermal equilibrium, is approximately 5.49°C. That’s quite a bit cooler than where we started! This problem beautifully illustrates the principles of calorimetry, heat transfer, and phase changes. Understanding these concepts is super useful for all sorts of applications, from engineering to everyday life.

Key Takeaways

• Calorimetry is about measuring heat exchange.
• Conservation of Energy is fundamental: heat lost = heat gained.
• Phase Changes (like melting) require energy.
• Specific Heat Capacity is crucial for calculating temperature changes.

Hope this breakdown helped you guys understand how to tackle these types of problems. Keep experimenting and stay curious!