Question

6.45 g of C6H12O6 is burned in a bomb calorimeter containing 950. g of water and the temperature goes from 35.0∘C to 42.3∘C. If the bomb has a heat capacity of 933 J∘C, what is the value for q of the reaction?

Answer 1

To calculate the heat of reaction (q), it is necessary to calculate the heat absorbed by the water and the bomb calorimeter individually, and then sum them for the total heat absorbed. Using the given mass of glucose, the temperature change, and the heat capacities of the bomb and the water, the total heat released during combustion of glucose in the calorimeter is found to be -36,212.9 J.

Step-by-step explanation:

To calculate the value for q of the reaction when glucose, C6H12O6, is burned in a bomb calorimeter, we must consider the heat absorbed by the water and the calorimeter bomb. The specific heat of water is 4.184 J/g°C. The total heat absorbed, q, can be found by using the formula:

q = -(qwater + qbomb)

Where:

• qwater = (mass of water) × (specific heat of water) × (change in temperature)
• qbomb = (heat capacity of bomb) × (change in temperature)

For the given problem:

• qwater = (950 g) × (4.184 J/g°C) × (42.3°C - 35.0°C)
• qbomb = (933 J/°C) × (42.3°C - 35.0°C)

q = -[(950 g × 4.184 J/g°C × 7.3°C) + (933 J/°C × 7.3°C)]

q = -[(29,402 J) + (6,810.9 J)] = -36,212.9 J

Therefore, the heat released (q) during the combustion of glucose in the bomb calorimeter is -36,212.9 J.

Answer 2

Answer: -35,800 J

Step-by-step explanation:

To solve this problem, first we an equation to relate the heat transfers involved.

qrxn+qwater+qbomb=0

This is is a consequence of the law of conservation of energy. In other words, this is true because heat can neither be created nor destroyed. Therefore, we solve for qrxn as follows.

qrxn=−[qwater+qbomb]

−[(4.184 Jg∘C)(950. g)(42.3∘C−35.0∘C)+(933 J∘C)(42.3∘C–35.0∘C)]

−(29,016 J+6,811 J)

−35,800 J