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: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=13.8 - 20.5=- 6.7^{\circ}C$

Step2: Use the heat - transfer formula

The formula for heat transfer is $Q = mc\Delta T$, where $m = 3.25g$, $c=0.129J/g^{\circ}C$, and $\Delta T=-6.7^{\circ}C$.
Substitute the values into the formula:
$Q=(3.25g)\times(0.129J/g^{\circ}C)\times(- 6.7^{\circ}C)$
First, calculate $3.25\times0.129 = 0.41925$
Then, $0.41925\times(-6.7)=-2.8099$
The negative sign indicates heat is released. Taking the magnitude, $Q = 2.81J$ (rounded to two decimal places)

Answer:

2.81 J