Answer to Question #169628 in General Chemistry for Samantha Hernandez

Question

Combustion of a fuel sample in a bomb calorimeter increases the temperature of the entire system by 5.10 °C if the calorimeter contains 1700 g of water, but only by 4.00 °C if the calorimeter contains 2200 g of water. What is the heat capacity of the dry bomb calorimeter assembly? Assume that the specific heat capacity of water is 4.18 J g^–1 °C^–1.

•  A. 570 J °C^–1
•  B. 912 J °C^–1
•  C. 366 J °C^–1
•  D. 722 J °C^–1
•  E. 494 J °C^–1

Answer 1

Q169628

Combustion of a fuel sample in a bomb calorimeter increases the temperature of the entire system by 5.10 °C if the calorimeter contains 1700 g of water, but only by 4.00 °C if the calorimeter contains 2200 g of water. What is the heat capacity of the dry bomb calorimeter assembly? Assume that the specific heat capacity of water is 4.18 J g^–1 °C^–1.

Solution:

Let us assume that the same assembly of bomb calorimeter and same amount of fuel is used in both the trials.

Since same amount of fuel us burned in both the trial, the amount of heat given out would be same too.

Heat absorbed by calorimeter and water in first trial is

Q ₁ = m ₁ * s * ΔT₁ + C_H * Δ T₁

where m₁ is the mass of water in first trial,

s is the specific heat capacity of water.

C_H is the heat capacity of calorimeter.

Δ T₁ is the change in temperature of calorimeter and water in first trial.

Heat absorbed by calorimeter and water in second trial is

Q ₂ = m ₂ * s * ΔT₂ + C_H * Δ T₂

where m₂ is the mass of water in first trial,

s is the specific heat capacity of water.

C_H is the heat capacity of calorimeter.

Δ T₂ is the change in temperature of calorimeter and water in second trial.

Amount of fuel used is same in both the trial, so amount of heat generated is also same.

Heat gained by calorimeter and water would also be same for both the trial.

That is , Q₁ = Q ₂

m ₁ * s * ΔT₁ + C_H * Δ T₁ = m ₂ * s * ΔT₂ + C_H * Δ T₂ ;

m₁ = 1700g, s = 4.18 J/g°C, Δ T₁ = 5.10 °C,

m ₂ = 2200g , Δ T₂ = 4.00 °C

substituting all this information we have

1700g * 4.18 J/g°C, * 5.10 °C + C_H * 5.10 °C = 2200g * 4.18 J/g°C, * 4.00 °C + C_H * 4.00 °C ;

36240.6 J + C_H * 5.10 °C = 36784 J + C_H * 4.00 °C

bring same terms on one side we have

C_H * 5.10 °C - C_H * 4.00 °C = 36784 J - 36240.6 J

C_H * ( 5.10 °C - 4.00 °C) = 543.4 J

C_H * 1.10 °C = 543.4 J ;

divide both the side by 1.10 °C

$\frac{ C_H \space * \space 1.10 °C }{ 1.10 °C } = \frac{543.4\space J }{1.10 \space °C}$

C_H = 494 J/ °C.

Hence the option E. 494 J °C^–1 is the correct option.