A 16.0- g
piece of glass at 65.2  ∘C
is dropped into 30.  g
of water at 23.3  ∘C
. All the heat lost by the glass is used to heat the water, and the final temperature of the water and glass is 25.9  ∘C
. Calculate the specific heat, in  J/g ∘C
, of the glass.

To calculate the specific heat (\(c\)) of the glass, we can use the principle of conservation of energy, which states that the heat lost by the glass is equal to the heat gained by the water. The formula to calculate heat (\(q\)) is:

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

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

For the glass:
- Mass (\(m_{\text{glass}}\)) = 16.0 g,
- Initial temperature (\(T_{\text{initial, glass}}\)) = 65.2 °C,
- Final temperature (\(T_{\text{final}}\)) = 25.9 °C,
- Specific heat (\(c_{\text{glass}}\)) = ?

For the water:
- Mass (\(m_{\text{water}}\)) = 30.0 g,
- Initial temperature (\(T_{\text{initial, water}}\)) = 23.3 °C,
- Final temperature (\(T_{\text{final}}\)) = 25.9 °C,
- Specific heat of water (\(c_{\text{water}}\)) = 4.18 J/g°C.

First, calculate the temperature changes:

\[
\Delta T_{\text{glass}} = T_{\text{final}} - T_{\text{initial, glass}} = 25.9 - 65.2 = -39.3 \, \text{°C}
\]

\[
\Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 25.9 - 23.3 = 2.6 \, \text{°C}
\]

Next, calculate the heat gained by the water:

\[
q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
\]
\[
q_{\text{water}} = 30.0 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 2.6 \, \text{°C}
\]
\[
q_{\text{water}} = 326.04 \, \text{J}
\]

Since the heat lost by the glass equals the heat gained by the water:

\[
q_{\text{glass}} = -q_{\text{water}} = -326.04 \, \text{J}
\]

Now, use the heat formula for the glass to solve for \(c_{\text{glass}}\):

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

\[
-326.04 \, \text{J} = 16.0 \, \text{g} \cdot c_{\text{glass}} \cdot (-39.3 \, \text{°C})
\]

Simplifying:

\[
c_{\text{glass}} = \frac{-326.04 \, \text{J}}{16.0 \, \text{g} \cdot (-39.3 \, \text{°C})}
\]

\[
c_{\text{glass}} \approx \frac{326.04 \, \text{J}}{628.8 \, \text{g°C}} \approx 0.5187 \, \text{J/g°C}
\]

So, the specific heat of the glass is approximately **0.519 J/g°C**.

Would you like more details, or do you have any questions?

Here are five related questions:
1. What is the specific heat capacity of common materials like iron or copper?
2. How does the specific heat capacity affect the rate of temperature change in materials?
3. Can you explain why the heat lost by the glass equals the heat gained by the water?
4. How would the final temperature change if the mass of the glass were different?
5. What would happen if the water and glass had the same initial temperature?

**Tip:** Always ensure units are consistent when performing calculations, especially when dealing with temperature, mass, and energy.

Learn how to calculate the specific heat of glass using principles of heat transfer and conservation of energy. This problem involves determining the specific heat capacity of glass from given temperatures and masses, transferring heat to water. Suitable for high school level understanding.

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