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Using the information from the table answer the questions 

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### Data Interpretation from the Table:
1. **Metal Cube:**
   - Mass = 144.64 g
   - Initial Temperature, \( T_i \) = 96.5°C
   - Final Temperature, \( T_f \) = 34.9°C
   - Change in Temperature, \( \Delta T = T_f - T_i = 34.9 - 96.5 = -61.6°C \)

2. **Water in Styrofoam:**
   - Volume = 55 mL (we will assume 1 mL of water weighs 1 g, so the mass of water = 55 g)
   - Initial Temperature, \( T_i \) = 21.8°C
   - Final Temperature, \( T_f \) = 34.9°C
   - Change in Temperature, \( \Delta T = T_f - T_i = 34.9 - 21.8 = 13.1°C \)

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### Questions and Answers:

#### 1. How much heat energy (\(q\)), in joules, was absorbed by the water?
We use the equation:
\[
q = m \cdot c \cdot \Delta T
\]
Where:
- \(m\) = mass of the water = 55 g,
- \(c\) = specific heat capacity of water = 4.184 J/g°C,
- \( \Delta T = 13.1°C \).

Thus:
\[
q = 55 \, \text{g} \times 4.184 \, \text{J/g°C} \times 13.1 \, \text{°C}
\]

Let's calculate this:

\[
q = 55 \times 4.184 \times 13.1 = 3016.84 \, \text{J} \approx 3000 \, \text{J}
\]

So, the water absorbed approximately **3000 J** of heat.

#### 2. What does the sign (+ or -) of \(q\) indicate?
The sign of \(q\) indicates whether heat was absorbed or released:
- A positive \(q\) means heat was **absorbed** by the system (in this case, water).
- A negative \(q\) would mean heat was **released**.

Since the water temperature increased, \(q\) is positive, indicating **heat absorption**.

#### 4. Solve for the specific heat capacity (\(c_p\)) of the unknown metal.
For the metal, we use the same formula for heat transfer but solve for \(c_p\):
\[
q = m \cdot c_p \cdot \Delta T
\]
Where:
- \( q = 3000 \, \text{J} \) (from water, assuming energy conservation),
- \( m = 144.64 \, \text{g} \),
- \( \Delta T = -61.6°C \) (temperature decrease of the metal).

Rearrange to solve for \(c_p\):
\[
c_p = \frac{q}{m \cdot \Delta T} = \frac{3000}{144.64 \times (-61.6)}
\]

Let me calculate that:

\[
c_p = \frac{3000}{144.64 \times 61.6} = \frac{3000}{8907.104} \approx 0.337 \, \text{J/g°C}
\]

So, the specific heat capacity of the metal is approximately **0.337 J/g°C**.

#### 5. Compare to the table of specific heat capacities.
From the table provided:
- Aluminum: \(0.897 \, \text{J/g°C}\)
- Brass: \(0.380 \, \text{J/g°C}\)
- Copper: \(0.385 \, \text{J/g°C}\)
- Carbon Steel: \(0.502 \, \text{J/g°C}\)

The calculated specific heat capacity (0.337 J/g°C) is closest to **Copper**, suggesting that the unknown metal is likely **Copper**.

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### Additional Questions:
1. How is the mass of water related to the calculation of heat absorbed?
2. Why is the specific heat capacity of water important in these calculations?
3. How does energy conservation play a role in calculating the heat transferred between the metal and water?
4. If the initial temperatures of the metal and water were different, how would that affect the final temperature and heat transfer?
5. What could cause discrepancies in the experimental vs. theoretical specific heat capacities?

**Tip:** Always check your units when performing specific heat calculations to ensure consistency and accuracy!

Using q = (m)(Cp)(ΔT) and your knowledge about heat transfer and the law of conservation of energy, calculate the amount of heat energy absorbed by water, determine the specific heat capacity of an unknown metal, and compare the experimental results with theoretical values.

Solve heat transfer problems by calculating the heat energy absorbed by water and the specific heat capacity of an unknown metal. Learn how to use q = m × Cp × ΔT and compare experimental values with theoretical results. Suitable for students in Grades 10-12.

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