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a reaction takes place in water. the waters initial temperature is 21.9…

Question

a reaction takes place in water. the waters initial temperature is 21.9 c and the final temperature is 76.2 c. the mass of the water is 61.2 g and the specific heat of water is 4.184 j/gc. calculate energy change (q) for this reaction.

Explanation:

Step1: Calculate the temperature change

The formula for temperature change $\Delta T$ is $\Delta T=T_{final}-T_{initial}$.
Given $T_{initial} = 21.9^{\circ}C$ and $T_{final}=76.2^{\circ}C$, so $\Delta T=76.2 - 21.9=54.3^{\circ}C$.

Step2: Use the heat - transfer formula

The heat - transfer formula is $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
Given $m = 61.2g$, $c = 4.184J/g^{\circ}C$, and $\Delta T = 54.3^{\circ}C$.
Substitute the values into the formula: $q=61.2\times4.184\times54.3$.
First, calculate $61.2\times4.184 = 256.0608$.
Then, $256.0608\times54.3=256.0608\times(50 + 4.3)=256.0608\times50+256.0608\times4.3$.
$256.0608\times50 = 12803.04$, $256.0608\times4.3=1101.06144$.
$q=12803.04+1101.06144 = 13904.10144\approx13904J$.

Answer:

$13904J$