Phase Changes - Free Printable

Problem Analysis:

The image provided is a graph titled "Temperature vs. Time," which plots the temperature of an object (in degrees Celsius) as a function of time (in minutes). The task likely involves analyzing the graph to extract information about the temperature changes over time. Since no specific question is stated, I will provide a general analysis of the graph and highlight key observations.

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Graph Description:

1. Axes:
- The horizontal axis represents Time (min), ranging from 0 to 26 minutes.
- The vertical axis represents Temperature (°C), ranging from 0°C to 160°C.

2. Temperature Changes:
- At time = 0 minutes, the temperature starts at approximately 160°C.
- From 0 to 2 minutes, the temperature decreases rapidly, dropping to around 120°C.
- From 2 to 10 minutes, the temperature remains relatively constant at 120°C.
- From 10 to 16 minutes, the temperature decreases linearly from 120°C to approximately 70°C.
- From 16 to 18 minutes, the temperature remains constant at around 70°C.
- From 18 to 24 minutes, the temperature decreases linearly from 70°C to approximately 20°C.
- At 24 minutes, the temperature reaches 20°C and remains constant until 26 minutes.

3. Key Observations:
- There are periods of rapid cooling (e.g., 0–2 minutes and 18–24 minutes).
- There are periods of constant temperature (e.g., 2–10 minutes and 16–18 minutes), indicating that the system may be reaching equilibrium or undergoing a phase change.
- The overall trend shows a continuous decrease in temperature over time, approaching room temperature (20°C).

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Possible Tasks and Solutions:

#### Task 1: Describe the cooling process.
- The object starts at a high temperature (160°C) and cools down over time. The cooling process is not uniform:
- Initially, there is rapid cooling (0–2 minutes).
- The temperature then stabilizes for a period (2–10 minutes), possibly due to a phase change or heat transfer equilibrium.
- Cooling resumes (10–16 minutes) before another stabilization period (16–18 minutes).
- Finally, the temperature drops rapidly again (18–24 minutes) until it reaches room temperature (20°C).

#### Task 2: Determine the rate of cooling during specific intervals.
- Interval 1: 0–2 minutes
- Initial temperature: 160°C
- Final temperature: 120°C
- Time elapsed: 2 minutes
- Rate of cooling:
\[
\text{Rate} = \frac{\Delta T}{\Delta t} = \frac{120 - 160}{2} = \frac{-40}{2} = -20 \, \text{°C/min}
\]
The negative sign indicates cooling.

- Interval 2: 10–16 minutes
- Initial temperature: 120°C
- Final temperature: 70°C
- Time elapsed: 6 minutes
- Rate of cooling:
\[
\text{Rate} = \frac{\Delta T}{\Delta t} = \frac{70 - 120}{6} = \frac{-50}{6} \approx -8.33 \, \text{°C/min}
\]

- Interval 3: 18–24 minutes
- Initial temperature: 70°C
- Final temperature: 20°C
- Time elapsed: 6 minutes
- Rate of cooling:
\[
\text{Rate} = \frac{\Delta T}{\Delta t} = \frac{20 - 70}{6} = \frac{-50}{6} \approx -8.33 \, \text{°C/min}
\]

#### Task 3: Identify periods of constant temperature.
- The temperature remains constant during two intervals:
- 2–10 minutes: Temperature is approximately 120°C.
- 16–18 minutes: Temperature is approximately 70°C.
- These periods suggest that the system may be undergoing phase changes (e.g., solidification or crystallization) where heat is being used to change the state of matter rather than lowering the temperature.

#### Task 4: Estimate the final temperature.
- The graph shows that the temperature approaches 20°C and remains constant after 24 minutes. This is likely the ambient room temperature.

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Final Answer:

Based on the analysis, the key points are:
1. The object cools rapidly initially, then stabilizes, cools again, stabilizes, and finally cools to room temperature.
2. The rate of cooling varies: \(-20 \, \text{°C/min}\) (0–2 minutes), \(-8.33 \, \text{°C/min}\) (10–16 minutes and 18–24 minutes).
3. Periods of constant temperature occur at 120°C (2–10 minutes) and 70°C (16–18 minutes).
4. The final temperature is 20°C.

\[
\boxed{20^\circ \text{C}}
\]