Question

2140 j of energy are released when 25.6 g of an unknown metal cools from 105.0 °c to 23.5 °c to achieve equilibrium.
$q_{h_2o} + q_{cal} = 2140$ j $q_{unknown} = -2140$ j
what is the specific heat capacity of the material?
$c = ?$ j/g °c

Explanation:

Step1: Recall the heat formula

The formula for heat energy is \( q = mc\Delta T \), where \( q \) is heat, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We know \( q_{\text{unknown}} = -2140 \, \text{J} \), \( m = 25.6 \, \text{g} \), and we need to find \( \Delta T \) first.
\( \Delta T = T_{\text{final}} - T_{\text{initial}} = 23.5^\circ\text{C} - 105.0^\circ\text{C} = -81.5^\circ\text{C} \)

Step2: Rearrange the formula to solve for \( c \)

From \( q = mc\Delta T \), we can rearrange to \( c = \frac{q}{m\Delta T} \). Substitute the known values: \( q = -2140 \, \text{J} \), \( m = 25.6 \, \text{g} \), \( \Delta T = -81.5^\circ\text{C} \).
\( c = \frac{-2140 \, \text{J}}{25.6 \, \text{g} \times (-81.5^\circ\text{C})} \)
First, calculate the denominator: \( 25.6 \times 81.5 = 25.6 \times 80 + 25.6 \times 1.5 = 2048 + 38.4 = 2086.4 \)
Then, \( c = \frac{2140}{2086.4} \approx 1.0256 \, \text{J/g}^\circ\text{C} \) (the negatives cancel out)

Answer:

\( \approx 1.03 \, \text{J/g}^\circ\text{C} \) (or more precisely \( \approx 1.026 \, \text{J/g}^\circ\text{C} \))