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_{cal}δt
c_{cal} = ? j/°c

Explanation:

Step1: Identify given values

We know that \( q_{rxn} = -843 \, \text{J} \) and \( \Delta T = 3.4 \, ^\circ\text{C} \). The formula relating these to the heat capacity of the calorimeter (\( c_{cal} \)) is \( -q_{rxn} = c_{cal} \Delta T \).

Step2: Rearrange the formula to solve for \( c_{cal} \)

We can rearrange the formula \( -q_{rxn} = c_{cal} \Delta T \) to solve for \( c_{cal} \) by dividing both sides by \( \Delta T \). This gives us \( c_{cal} = \frac{-q_{rxn}}{\Delta T} \).

Step3: Substitute the given values into the formula

Substitute \( q_{rxn} = -843 \, \text{J} \) and \( \Delta T = 3.4 \, ^\circ\text{C} \) into the formula. First, calculate \( -q_{rxn} \): \( -(-843 \, \text{J}) = 843 \, \text{J} \). Then, divide this by \( \Delta T \): \( c_{cal} = \frac{843 \, \text{J}}{3.4 \, ^\circ\text{C}} \).

Step4: Perform the division

Calculating \( \frac{843}{3.4} \approx 247.94 \). So, \( c_{cal} \approx 248 \, \text{J/}^\circ\text{C} \) (rounded to a reasonable number of significant figures).

Answer:

\( \boxed{248} \) (or approximately \( 247.9 \) depending on rounding, but \( 248 \) is a common result here)