Answer:
[tex]\boxed {\boxed {\sf 39,668.2 \ Joules}}[/tex]
Explanation:
We are given the mass and change in temperature, so we must use this formula for heat energy:
[tex]q=mc \Delta T[/tex]
The mass is 73 grams. Water's specific heat is 4.18 J/g × °C. Let's calculate the change in temperature
- ΔT= final temperature - initial temperature
- ΔT= 43 °C - 30°C
- ΔT= 13 °C
Now we know all the variables and can substitute them into the formula.
[tex]m= 73 \ g \\c= 4.18 \ J/g* \textdegree C \\\Delta T= 13 \ \textdegree C[/tex]
[tex]q= (73 \ g )(4.18 \ J/g*\textdegree C)(13 \textdegree C)[/tex]
Multiply the first numbers together. The grams will cancel.
[tex]q= 3051.4 \ J/\textdegree C (13 \textdegree C)[/tex]
Multiply again. This time, the degrees Celsius cancel.
[tex]q= 39668.2 \ J[/tex]
39,668.2 Joules of heat energy are absorbed.
