Question

an electric heater supplies 46.9 joules of energy to 13.5-g sample of ch3co2h originally at 18.8°c. compute the final temperature (in °c).

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 capacity, and \( \Delta T = T_f - T_i \) (final temperature - initial temperature). For acetic acid (\( \text{CH}_3\text{CO}_2\text{H} \)), the specific heat capacity \( c = 2.04 \, \text{J/g}^\circ\text{C} \) (a known value for acetic acid). We know \( q = 46.9 \, \text{J} \), \( m = 13.5 \, \text{g} \), and \( T_i = 18.8^\circ\text{C} \). We need to find \( T_f \).

First, rearrange the formula to solve for \( \Delta T \): \( \Delta T = \frac{q}{mc} \)

Step2: Calculate \( \Delta T \)

Substitute the known values into the formula:
\( \Delta T = \frac{46.9 \, \text{J}}{13.5 \, \text{g} \times 2.04 \, \text{J/g}^\circ\text{C}} \)
First, calculate the denominator: \( 13.5 \times 2.04 = 27.54 \)
Then, \( \Delta T = \frac{46.9}{27.54} \approx 1.703^\circ\text{C} \)

Step3: Calculate the final temperature \( T_f \)

We know that \( \Delta T = T_f - T_i \), so \( T_f = T_i + \Delta T \)
Substitute \( T_i = 18.8^\circ\text{C} \) and \( \Delta T \approx 1.703^\circ\text{C} \):
\( T_f = 18.8 + 1.703 \approx 20.5^\circ\text{C} \) (rounded to a reasonable number of significant figures)

Answer:

\( 20.5 \) (or a more precise value depending on calculation, but approximately \( 20.5^\circ\text{C} \))