Question

the combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4 °c.
q_rxn = -843 j
what is the heat capacity of the calorimeter?
-q_rxn = c_calδt
c_cal = ? j/°c

Explanation:

Step1: Recall the formula

We have the formula \(-q_{rxn} = c_{cal}\Delta T\), and we need to solve for \(c_{cal}\). So we can rearrange the formula to \(c_{cal}=\frac{-q_{rxn}}{\Delta T}\).

Step2: Substitute the values

We know that \(q_{rxn}=-843\space J\) and \(\Delta T = 3.4\space^{\circ}C\). Substituting these values into the formula:
\(c_{cal}=\frac{-(-843\space J)}{3.4\space^{\circ}C}=\frac{843\space J}{3.4\space^{\circ}C}\)

Step3: Calculate the result

\(\frac{843}{3.4}\approx247.94\space J/^{\circ}C\) (rounded to a reasonable number of significant figures)

Answer:

\(248\) (or approximately \(247.9\)) \(J/^{\circ}C\) (If we calculate more precisely \(\frac{843}{3.4}=247.941176\cdots\), so we can present it as \(248\) or \(247.9\) depending on the required precision)