a hydrocarbon reacts with oxygen in a bomb calorimeter containing 2.00 kg of water. the temperature rises by…

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?\nq_cal = q_h₂o + q_dry\nq_cal = ? j\nenter either a + or - sign and the magnitude.\ndo not round.

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?\nq_cal = q_h₂o + q_dry\nq_cal = ? j\nenter either a + or - sign and the magnitude.\ndo not round.

Answer

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