Questions: A 55.4 g sample of glass, which has a specific heat capacity of 0.670 J ⋅ g^-1 ⋅ °C^-1, is put into a calorimeter that contains 300.0 g of water. The temperature of the water starts off at 21.0°C. When the temperature of the water stops changing it's 23.1°C. The pressure remains constant at 1 atm. Calculate the initial temperature of the glass sample. Be sure your answer is rounded to 2 significant digits.

Solution Steps

We need to find the initial temperature of the glass sample. We know:

• Mass of glass, \( m_{\text{glass}} = 55.4 \, \text{g} \)
• Specific heat capacity of glass, \( c_{\text{glass}} = 0.670 \, \text{J/g} \cdot {}^{\circ} \text{C} \)
• Mass of water, \( m_{\text{water}} = 300.0 \, \text{g} \)
• Specific heat capacity of water, \( c_{\text{water}} = 4.184 \, \text{J/g} \cdot {}^{\circ} \text{C} \)
• Initial temperature of water, \( T_{\text{water, initial}} = 21.0^{\circ} \text{C} \)
• Final temperature of water and glass, \( T_{\text{final}} = 23.1^{\circ} \text{C} \)

The heat lost by the glass is equal to the heat gained by the water. Therefore, we can write:

\[ m_{\text{glass}} \cdot c_{\text{glass}} \cdot (T_{\text{glass, initial}} - T_{\text{final}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{water, initial}}) \]

Substitute the known values into the equation:

\[ 55.4 \cdot 0.670 \cdot (T_{\text{glass, initial}} - 23.1) = 300.0 \cdot 4.184 \cdot (23.1 - 21.0) \]

Calculate the right side:

\[ 300.0 \cdot 4.184 \cdot 2.1 = 2631.12 \, \text{J} \]

Now solve for \( T_{\text{glass, initial}} \):

\[ 37.118 \cdot (T_{\text{glass, initial}} - 23.1) = 2631.12 \]

\[ T_{\text{glass, initial}} - 23.1 = \frac{2631.12}{37.118} \]

\[ T_{\text{glass, initial}} - 23.1 = 70.88 \]

\[ T_{\text{glass, initial}} = 70.88 + 23.1 = 93.98 \]

Final Answer