question 9 (1 point)\na 3.25 g piece of gold is cooled from 20.5°c to 13.8°c. the specific heat capacity of…

question 9 (1 point)\na 3.25 g piece of gold is cooled from 20.5°c to 13.8°c. the specific heat capacity of gold is\n0.129 j/g°c. how much heat is released?\n2.81 j\n2.80 j\n2.8 j\n5.78 j\n8.59 j

question 9 (1 point)\na 3.25 g piece of gold is cooled from 20.5°c to 13.8°c. the specific heat capacity of gold is\n0.129 j/g°c. how much heat is released?\n2.81 j\n2.80 j\n2.8 j\n5.78 j\n8.59 j

Answer

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