Question

a hydrocarbon reacts with oxygen in a bomb calorimeter containing 2.00 kg of water. the temperature rises by 0.658 °c. the heat capacity of the dry portion of the calorimeter is 1050 j/°c. what is the heat of the overall calorimeter in joules?
q_cal = q_h₂o + q_dry
q_cal = ? j
enter either a + or - sign and the magnitude.
do not round.

Explanation:

Step1: Calculate \( q_{H_2O} \)

The formula for heat absorbed by water is \( q = mc\Delta T \). The mass of water \( m = 2.00\ kg = 2000\ g \), specific heat of water \( c = 4.184\ J/g^\circ C \), and \( \Delta T = 0.658^\circ C \).
So, \( q_{H_2O}=2000\ g\times4.184\ J/g^\circ C\times0.658^\circ C \)
\( q_{H_2O}=2000\times4.184\times0.658 \)
\( q_{H_2O}=5493.664\ J \)

Step2: Calculate \( q_{dry} \)

The formula for heat absorbed by the dry portion is \( q = C\Delta T \), where \( C = 1050\ J/^\circ C \) and \( \Delta T = 0.658^\circ C \).
So, \( q_{dry}=1050\ J/^\circ C\times0.658^\circ C = 690.9\ J \)

Step3: Calculate \( q_{cal} \)

Using \( q_{cal}=q_{H_2O}+q_{dry} \)
\( q_{cal}=5493.664\ J + 690.9\ J \)
\( q_{cal}=6184.564\ J \)
Since the reaction is exothermic (hydrocarbon burning), the calorimeter absorbs heat, so \( q_{cal} \) is positive.

Answer:

+6184.564