Question

a 20.3 g sample of an unknown metal and a 28.5 g sample of copper, both at 80.6 °c, are added to 104. g of water at 11.2 °c in a constant-pressure calorimeter of negligible heat capacity. if the final temperature of the metals and water is 16.1 °c, determine the specific heat of the unknown metal. be sure your answer has the correct number of significant digits. note: reference the phase change properties of pure substances table for additional information.

Explanation:

Step1: Recall heat transfer principle

Heat lost by metals = Heat gained by water. Formula: \( q_{\text{lost}} = -q_{\text{gained}} \), where \( q = mc\Delta T \), \( m \) = mass, \( c \) = specific heat, \( \Delta T = T_{\text{final}} - T_{\text{initial}} \).

Step2: Identify known values

• For water: \( m_w = 104. \, \text{g} \), \( c_w = 4.184 \, \frac{\text{J}}{\text{g·°C}} \), \( T_{i,w} = 11.2 \, \text{°C} \), \( T_f = 16.1 \, \text{°C} \)
• For copper: \( m_{Cu} = 28.5 \, \text{g} \), \( c_{Cu} = 0.385 \, \frac{\text{J}}{\text{g·°C}} \), \( T_{i,Cu} = 80.6 \, \text{°C} \)
• For unknown metal: \( m_{unk} = 20.3 \, \text{g} \), \( T_{i,unk} = 80.6 \, \text{°C} \), \( c_{unk} = ? \)

Step3: Calculate heat gained by water

\( \Delta T_w = 16.1 - 11.2 = 4.9 \, \text{°C} \)
\( q_w = m_w c_w \Delta T_w = 104. \times 4.184 \times 4.9 \)
\( q_w = 104. \times 20.4016 \approx 2121.7664 \, \text{J} \)

Step4: Calculate heat lost by copper

\( \Delta T_{Cu} = 16.1 - 80.6 = -64.5 \, \text{°C} \)
\( q_{Cu} = m_{Cu} c_{Cu} \Delta T_{Cu} = 28.5 \times 0.385 \times (-64.5) \)
\( q_{Cu} = 28.5 \times (-24.8325) \approx -707.72625 \, \text{J} \)

Step5: Calculate heat lost by unknown metal

Let \( q_{unk} = m_{unk} c_{unk} \Delta T_{unk} \), \( \Delta T_{unk} = 16.1 - 80.6 = -64.5 \, \text{°C} \)
Total heat lost: \( q_{lost} = q_{Cu} + q_{unk} = -q_w \)
\( -707.72625 + 20.3 \times c_{unk} \times (-64.5) = -2121.7664 \)
Simplify: \( -707.72625 - 1309.35 c_{unk} = -2121.7664 \)
Add 707.72625 to both sides: \( -1309.35 c_{unk} = -2121.7664 + 707.72625 = -1414.04015 \)
Divide by -1309.35: \( c_{unk} = \frac{1414.04015}{1309.35} \approx 1.08 \, \frac{\text{J}}{\text{g·°C}} \)

Answer:

\( \boxed{1.08} \) (in \( \frac{\text{J}}{\text{g·°C}} \))