Please I need your help in solving this. = A 0.3423g sample of pentane c5H12 was burned in a bomb calorimeter, the temperature of the calorimeter and the 1.00kg of water contained therein rose from 20.22°c to 22.82 °c. The heat capacity of the calorimeter is 2.21kj/°c. The heat capacity of water is 4.184j/g°c. How much heat was given off during combustion of the sample of pentane?
To find the amount of heat given off during the combustion of the sample of pentane, you can use the formula:
q = mcΔT
where:
q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
First, let's calculate the heat energy absorbed by the water:
q_water = mcΔT
q_water = (1000 g) * (4.184 J/g°C) * (22.82 °C - 20.22 °C)
Calculating the value of q_water gives:
q_water ≈ (1000 g) * (4.184 J/g°C) * (2.6 °C)
Next, let's calculate the heat energy absorbed by the calorimeter:
q_calorimeter = mcΔT
q_calorimeter = (2.21 kJ/°C) * (22.82 °C - 20.22 °C)
Calculating the value of q_calorimeter gives:
q_calorimeter ≈ (2.21 kJ/°C) * (2.6 °C)
Finally, to find the total heat energy given off during the combustion of the sample of pentane, we add the values of q_water and q_calorimeter:
q_total = q_water + q_calorimeter
Substituting the calculated values:
q_total ≈ (1000 g) * (4.184 J/g°C) * (2.6 °C) + (2.21 kJ/°C) * (2.6 °C)
Evaluating the expression gives:
q_total ≈ 10,372.8 J + 5726 J
Therefore, the heat given off during the combustion of the sample is approximately 16,098.8 J.
To solve this problem, we can use the equation:
q = m × c × ΔT
Where:
q is the heat energy (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
First, let's calculate the heat energy released by the water.
Given:
Mass of water (m) = 1000 g (1.00 kg)
Specific heat capacity of water (c) = 4.184 J/g°C
Change in temperature (ΔT) = 22.82°C - 20.22°C = 2.60°C
q_water = m × c × ΔT = 1000 g × 4.184 J/g°C × 2.60°C
q_water ≈ 10,835.36 J
Next, let's calculate the heat energy absorbed by the calorimeter.
Given:
Heat capacity of the calorimeter (C_cal) = 2.21 kJ/°C = 2,210 J/°C
Change in temperature (ΔT) = 22.82°C - 20.22°C = 2.60°C
q_cal = C_cal × ΔT = 2,210 J/°C × 2.60°C
q_cal ≈ 5,746 J
Now, let's calculate the total heat energy released during combustion.
q_total = q_water + q_cal
q_total ≈ 10,835.36 J + 5,746 J
q_total ≈ 16,581.36 J
Therefore, approximately 16,581.36 J of heat was given off during the combustion of the sample of pentane.
This a multpart question usually. You have asked just one part of it. When you have multipart questions it is best to post the entire question and let us see the whole thing instead of post each part at separate times.
q = total heat = [mass H2O x heat capacity H2O x (Tfinal-Tinitial)] | + [heat capacity calorimeter x (Tfinal-Tinitial)]