Question

a 5.25 g sample of metal gives off 10.4 j of energy as it cools from 49.5 °c to 40.5 °c. what is the specific heat of the metal?
$c = ? \frac{j}{g \cdot ^\circ c}$
note: $q = -10.4 j$

Explanation:

Step1: Recall the formula for heat transfer

The formula for heat transfer is \( q = mc\Delta T \), where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature. We need to solve for \( c \), so we can rearrange the formula to \( c=\frac{q}{m\Delta T} \).

Step2: Calculate the change in temperature (\(\Delta T\))

The initial temperature \( T_i = 49.5^\circ C \) and the final temperature \( T_f = 40.5^\circ C \). So, \( \Delta T=T_f - T_i = 40.5^\circ C - 49.5^\circ C=- 9.0^\circ C \). The magnitude of the temperature change is what matters for the calculation (since we're dealing with energy loss, the sign of \( q \) already accounts for the direction).

Step3: Substitute the known values into the formula for \( c \)

We know that \( q = - 10.4\space J \), \( m = 5.25\space g \), and \( \Delta T=-9.0^\circ C \). Substituting into \( c=\frac{q}{m\Delta T} \), we get \( c=\frac{- 10.4\space J}{5.25\space g\times(-9.0^\circ C)} \).

Step4: Perform the calculation

First, calculate the denominator: \( 5.25\space g\times9.0^\circ C = 47.25\space g\cdot^\circ C \) (the negatives cancel out). Then, divide the numerator by the denominator: \( c=\frac{10.4\space J}{47.25\space g\cdot^\circ C}\approx0.220\space\frac{J}{g\cdot^\circ C} \)

Answer:

\( \boldsymbol{0.220} \) (or approximately \( 0.22 \)) \( \frac{J}{g\cdot^\circ C} \)