Respuesta :
Answer:
The heat capacity of the calorimeter is 4.76 kJ/°C
Explanation:
Step 1: Data given
1.00 mol of glucose releases 2820 kJ of heat
Mass of glucose = 2.0 grams
Mass of water = 1000 grams
The temperature increases with 3.5 °C
Step 2: Calculate moles
moles glucose = mass glucose / molar mass glucose
moles glucose = 2.0 grams / 180.16 g/mol
moles glucose = 0.0111 moles
Step 3: Calculate heat produced by the combustion
Heat produced = 2820 kJ/mol * 0.0111 moles
Heat produced = 31.302 kJ = 31302 J
Step 4: Calculate heat absorbed by the water
Q = m*c*ΔT
⇒ with m = the mass of water = 1000 grams
⇒ with c = the specific heat of water = 4.184 J/g°C
⇒ with ΔT = The change in temperature = 3.5 °C
Q = 1000 * 4.184 *3.5
Q = 14644 J absorbed by the water
Step 5: Calculate heat basorbed by the calorimeter
Q = 31302 - 14644 = 16658 J absorbed by the calorimeter
Step 6: Calculate the heat capacity of the calorimeter
c= 16658 J / 3.5 °C
c = 4759 J/°C = 4.76 kJ/°C
The heat capacity of the calorimeter is 4.76 kJ/°C
The specific heat capacity of the calorimeter is 4.759 kJ/⁰C.
The molar mass of glucose is calculated as;
C₆H₁₂O₆ = (12 x 6) + (1 x 12) + (16 x 6) = 180 g/mol
The number of moles in 2g of glucose;
[tex]mole = \frac{2}{180} \\\\mole = 0.0111 \ mol[/tex]
The heat added by 0.011 mol of glucose
1 mole -------------- 2820 kJ
0.0111 mole --------- ?
[tex]= 0.0111 \times 2820 \ kJ\\\\= 31.302 \ kJ\\\\= 31,302 \ J[/tex]
The heat of the given water;
[tex]Q = mc\Delta \theta \\\\Q = (1000)(4.184)(3.5)\\\\Q = 14,644 \ J[/tex]
The heat absorbed by the calorimeter is calculated as follows;
[tex]\Delta Q = 31,302\ J - 14.644 \ J\\\\\Delta Q = 16,658 \ J[/tex]
The specific heat capacity of the calorimeter is calculated as follows;
[tex]C_c = \frac{\Delta Q }{\Delta \theta} \\\\C_c = \frac{16,658}{3.5} \\\\C_c = 4.759 \ kJ/^0C[/tex]
Learn more here:https://brainly.com/question/16990783
