Question

question 9 (1 point)
a 3.25 g piece of gold is cooled from 20.5°c to 13.8°c. the specific heat capacity of gold is
0.129 j/g°c. how much heat is released?
2.81 j
2.80 j
2.8 j
5.78 j
8.59 j

Explanation:

Step1: Recall the heat formula

The formula for heat \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Calculate the change in temperature

\( \Delta T = T_{final} - T_{initial} = 13.8^\circ\text{C} - 20.5^\circ\text{C} = -6.7^\circ\text{C} \). The negative sign indicates heat is released.

Step3: Substitute values into the formula

\( m = 3.25 \, \text{g} \), \( c = 0.129 \, \text{J/g}^\circ\text{C} \), \( \Delta T = -6.7^\circ\text{C} \).
\( q = 3.25 \times 0.129 \times (-6.7) \). First, calculate \( 3.25 \times 0.129 = 0.41925 \). Then, \( 0.41925 \times (-6.7) \approx -2.81 \, \text{J} \). The magnitude is 2.81 J (heat released).

Answer:

2.81 J