A piece of unknown metal with a mass of 25.0 grams is placed in calorimeter.  The water in the calorimeter initially is at 24.0C and the metal’s initial temperature is 99.0C.  The metal and the water in the calorimeter come to a final temperature of 26.5C.  If the specific heat of water is 4.184 J/g(C) and there was 350 grams of water in the calorimeter, what is the specific heat of the unknown metal?

This is a classic calorimetry problem where the heat lost by the metal is equal to the heat gained by the water. We can solve this problem using the formula for heat transfer:

\[
q = m \cdot c \cdot \Delta T
\]

Where:
- \( q \) is the heat transferred,
- \( m \) is the mass,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

### Step 1: Calculate the heat gained by the water

The heat gained by the water can be calculated using the formula:

\[
q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
\]

Where:
- \( m_{\text{water}} = 350 \, \text{g} \),
- \( c_{\text{water}} = 4.184 \, \text{J/g(°C)} \),
- \( \Delta T_{\text{water}} = 26.5 - 24.0 = 2.5 \, \text{°C} \).

So,

\[
q_{\text{water}} = 350 \cdot 4.184 \cdot 2.5
\]

### Step 2: Calculate the heat lost by the metal

The heat lost by the metal is equal to the heat gained by the water (since energy is conserved), but we can also express this heat as:

\[
q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
\]

Where:
- \( m_{\text{metal}} = 25.0 \, \text{g} \),
- \( \Delta T_{\text{metal}} = 99.0 - 26.5 = 72.5 \, \text{°C} \),
- \( c_{\text{metal}} \) is the unknown.

Since \( q_{\text{metal}} = -q_{\text{water}} \) (because the metal loses heat while the water gains heat), we can solve for \( c_{\text{metal}} \) once we find \( q_{\text{water}} \).

Let's calculate both \( q_{\text{water}} \) and \( c_{\text{metal}} \).The heat gained by the water is \( 3661.0 \, \text{J} \). Since the heat lost by the metal is equal to the heat gained by the water (but with opposite signs), the specific heat capacity of the unknown metal is approximately:

\[
c_{\text{metal}} = 2.02 \, \text{J/g(°C)}
\]

### Do you have any further questions or would you like more details?

Here are 5 related questions to further explore the topic:
1. How would the result change if the mass of the water was doubled?
2. What would happen if the final temperature of the system was higher than 26.5°C?
3. How does the specific heat of the unknown metal compare to common metals like copper or aluminum?
4. Can you determine the identity of the metal based on the specific heat value?
5. What assumptions are made in this calorimetry experiment?

**Tip:** Always ensure that the calorimeter used is well-insulated to minimize heat loss to the surroundings for more accurate results.

A piece of unknown metal with a mass of 25.0 grams is placed in a calorimeter. The water in the calorimeter initially is at 24.0°C and the metal’s initial temperature is 99.0°C. The metal and the water in the calorimeter come to a final temperature of 26.5°C. If the specific heat of water is 4.184 J/g(°C) and there was 350 grams of water in the calorimeter, what is the specific heat of the unknown metal?

Solve this calorimetry problem to determine the specific heat of an unknown metal. A 25.0 g metal sample is placed in a calorimeter with 350 g of water, initially at 24.0°C, and comes to a final temperature of 26.5°C. This problem explores key thermodynamic principles, including the formula q = m * c * ΔT and the Law of Conservation of Energy. Suitable for Grades 10-12.

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