When a 1.010 g sample of sucrose (MW = 342.30 g/mol)

When a 1.010 g sample of sucrose (MW = 342.30 g/mol)

 

 

When a 1.010 g sample of sucrose (MW = 342.30 g/mol) is completely combusted in a bomb calorimeter, the temperature of the calorimeter is increased by 4.50 ℃. If the heat capacity of calorimeter is 3.75 kJ/℃, how much heat was absorbed by the calorimeter?

 

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열용량의 정의

C = q / Δt

( 식 설명 https://ywpop.tistory.com/2897 )

 

 

q = C Δt

= (3.75 kJ/℃) (4.50 ℃)

= 16.875 kJ

 

 

답: 16.9 kJ

 

 

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The combustion of 1.010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24.92 to 28.33 ℃. The heat capacity of the calorimeter assembly is 4.90 kJ/℃.

(a) What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11?

(b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

 

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q_cal = C Δt

= (4.90 kJ/℃) (28.33 – 24.92 ℃)

= 16.709 kJ

 

 

q_rxn = -16.709 kJ / 1.010 g

= -16.5 kJ/g

 

 

또는 C12H22O11의 몰질량 = 342.30 g/mol 이므로,

= (-16.709 kJ / 1.010 g) × (342.30 g/mol)

= -5662.9 kJ/mol

= -5.66×10^3 kJ/mol

 

 

Calculate q_rxn for one teaspoon:

(-16.5 kJ/g) × (4.8 g / one teaspoon) × (1 kcal / 4.184 kJ)

= -18.9 kcal / one teaspoon