Question

use this data to find q for the experiment. specific heat 4.184 j/gc initial temperature 26.3 final temperature 75.9 mass of water 86.82

Explanation:

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=75.9 - 26.3=49.6^{\circ}C$

Step2: Use heat formula

The formula for heat $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is temperature change.
Substitute $m = 86.82g$, $c=4.184J/g^{\circ}C$, $\Delta T = 49.6^{\circ}C$ into the formula:
$q=(86.82)(4.184)(49.6)$
First calculate $(86.82)(4.184)=86.82\times4.184 = 363.26448$
Then $q=363.26448\times49.6\approx18018.92J$

Answer:

$18018.92J$