Question

a 50.00 g sample of an unknown metal is heated to 45.00°c. it is then placed in a coffee-cup calorimeter filled with water. the calorimeter and the water have a combined mass of 250.0 g and an overall specific heat of 1.035 cal/g·°c. the initial temperature of the calorimeter is 10.00°c. the system reaches a final temperature of 11.08°c when the metal is added. which of these equations is the correct one to use to calculate the specific heat of the metal?
$c_{p, \text{metal}} = \frac{-m_{\text{cal}}c_{p, \text{cal}}\delta t_{\text{cal}}}{m_{\text{metal}}\delta t_{\text{metal}}}$
$c_{p, \text{metal}} = \frac{m_{\text{metal}}\delta t_{\text{metal}}}{m_{\text{cal}}c_{p, \text{cal}}\delta t_{\text{cal}}}$
what is the specific heat of the metal?
\boxed{} cal/g·°c

Explanation:

Part 1: Correct Equation for Specific Heat of Metal

In calorimetry, the heat lost by the metal ($q_{\text{metal}}$) is equal to the heat gained by the water and calorimeter ($q_{\text{cal}}$), but with opposite signs (since heat lost is negative and heat gained is positive). The formula for heat is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is change in temperature. So, $q_{\text{metal}} = -q_{\text{cal}}$. Substituting the heat formulas: $m_{\text{metal}}c_{\text{p,metal}}\Delta T_{\text{metal}} = -m_{\text{cal}}c_{\text{p,cal}}\Delta T_{\text{cal}}$. Solving for $c_{\text{p,metal}}$ gives $c_{\text{p,metal}} = \frac{-m_{\text{cal}}c_{\text{p,cal}}\Delta T_{\text{cal}}}{m_{\text{metal}}\Delta T_{\text{metal}}}$, which matches the first equation.

Step 1: Identify Values

• $m_{\text{metal}} = 50.00\ \text{g}$
• $T_{\text{initial,metal}} = 45.00^\circ\text{C}$, $T_{\text{final,metal}} = 11.08^\circ\text{C}$, so $\Delta T_{\text{metal}} = 11.08 - 45.00 = -33.92^\circ\text{C}$
• $m_{\text{cal}} = 250.0\ \text{g}$, $c_{\text{p,cal}} = 1.035\ \text{cal/g}\cdot^\circ\text{C}$
• $T_{\text{initial,cal}} = 10.00^\circ\text{C}$, $T_{\text{final,cal}} = 11.08^\circ\text{C}$, so $\Delta T_{\text{cal}} = 11.08 - 10.00 = 1.08^\circ\text{C}$

Step 2: Substitute into the Correct Formula

Using $c_{\text{p,metal}} = \frac{-m_{\text{cal}}c_{\text{p,cal}}\Delta T_{\text{cal}}}{m_{\text{metal}}\Delta T_{\text{metal}}}$:

$$ c_{\text{p,metal}} = \frac{-(250.0\ \text{g})(1.035\ \text{cal/g}\cdot^\circ\text{C})(1.08^\circ\text{C})}{(50.00\ \text{g})(-33.92^\circ\text{C})} $$

Step 3: Calculate Numerator and Denominator

• Numerator: $-(250.0)(1.035)(1.08) = -(250.0 \times 1.035 \times 1.08) = -(284.55)$ (the negative signs will cancel)
• Denominator: $(50.00)(-33.92) = -1696$

Step 4: Divide to Find $c_{\text{p,metal}}$

$$ c_{\text{p,metal}} = \frac{284.55}{1696} \approx 0.1678\ \text{cal/g}\cdot^\circ\text{C} $$

Answer:

The correct equation is $C_{\text{p,metal}} = \frac{-m_{\text{cal}}C_{\text{p,cal}}\Delta T_{\text{cal}}}{m_{\text{metal}}\Delta T_{\text{metal}}}$ (the first given equation).

Part 2: Calculating Specific Heat of the Metal