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 heat formula

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

Step2: Calculate \(\Delta T\)

\(\Delta T = T_{final}-T_{initial}=40.5^\circ C - 49.5^\circ C=-9.0^\circ C\). The negative sign indicates cooling.

Step3: Substitute values into the formula

We know \( q = -10.4\space J \), \( m = 5.25\space g \), and \( \Delta T = -9.0^\circ C \). Plugging 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). Then, \( c=\frac{10.4\space J}{47.25\space g\cdot^\circ C}\approx0.220\space\frac{J}{g\cdot^\circ C} \).

Answer:

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