A 1.55-g sample of propane, C3H8, was burned in a bomb calorimeter whose total heat capacity is 12.3 kJ/°C. The temperature of the calorimeter plus its contents increased from 21.36°C to 27.69°C. What is the heat of combustion per gram of propane?

50.23

To solve this problem, we need to use the equation:

\(q = mc\Delta T\)

Where:
- \(q\) is the heat absorbed or released,
- \(m\) is the mass of the substance, and
- \(c\) is the specific heat capacity.

First, let's find the heat absorbed by the calorimeter and its contents. We can use the equation above and the given information. The mass of the calorimeter and its contents is not given, but we will assume it remains constant throughout the experiment.

\(\Delta T = T_f - T_i\)

\(\Delta T = 27.69°C - 21.36°C\)

\(\Delta T = 6.33°C\)

Next, we need to convert the mass of propane from grams to kilograms to match the units of the specific heat capacity.

\(m = 1.55 g \times \frac{1 kg}{1000 g}\)

\(m = 0.00155 kg\)

Now we can calculate the heat absorbed.

\(q = mc\Delta T\)

\(q = 0.00155 kg \times 12.3 kJ/°C \times 6.33°C\)

\(q = 0.1190785 kJ\)

Finally, to find the heat of combustion per gram of propane, we can divide the heat absorbed by the mass of propane.

Heat of Combustion per gram of propane = \(\frac{q}{m}\)

Heat of Combustion per gram of propane = \(\frac{0.1190785 kJ}{1.55 g}\)

Heat of Combustion per gram of propane ≈ 0.076881 kJ/g

Therefore, the heat of combustion per gram of propane is approximately 0.076881 kJ/g.

To find the heat of combustion per gram of propane, we need to calculate the amount of heat released by burning the given mass of propane and then divide it by the mass of the propane.

First, let's find the heat released during the combustion of the given mass of propane. We can use the formula:

Heat released = (Change in temperature) x (Heat capacity of the calorimeter)

The change in temperature is given as 27.69°C - 21.36°C = 6.33°C.

The heat capacity of the calorimeter is given as 12.3 kJ/°C.

Converting the heat capacity from kJ/°C to J/°C (since the mass of propane is given in grams), we have:

Heat capacity = 12.3 kJ/°C x 1000 J/1 kJ = 12,300 J/°C.

Now, let's calculate the heat released:

Heat released = 6.33°C x 12,300 J/°C = 77,799 J.

The given mass of propane is 1.55 g.

Finally, we can calculate the heat of combustion per gram of propane:

Heat of combustion = Heat released / Mass of propane
= 77,799 J / 1.55 g
= 50,193.55 J/g

Therefore, the heat of combustion per gram of propane is approximately 50,193.55 J/g.

I worked a quite similar problem on an hour or so ago. The same process applies.