Suppose that coal of density 1.5 g/cm3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first approximation.) The combustion of carbon is described by the equation C(s) + O2(g) −! CO2(g) �H◦ = −394 kJ
What is the value of q (heat) when a lump of coal of size 8 cm×7.5 cm×7 cm is burned?
Answer in units of kJ
The next question follows the first one
:
What mass of water can be heated from 27◦C to 100◦C by burning this piece of coal?
Answer in units of g
For the first question, the answer I got was -20665.3 kJ and it was correct. However, every time I plug that answer into the equation I get the wrong answer for the second question. I've converted the aforementioned number into joules but I still got the wrong answer when I plugged it into the equation. I'm not sure what I'm doing wrong.
Well, well, well, let's burn some coal and heat things up! *cue evil laughter*
To calculate the heat (q) released when burning coal, we need to first find the volume of the lump of coal. The volume can be calculated by multiplying the length, width, and height of the coal.
Volume = 8 cm × 7.5 cm × 7 cm = 420 cm³
Since the density of coal is given as 1.5 g/cm³, we can find the mass of the coal by multiplying the volume by the density.
Mass = Volume × Density = 420 cm³ × 1.5 g/cm³ = 630 g
Now, we know the mass of the coal, but the heat released is given per mole of carbon. So we need to convert the mass of coal to moles of carbon using the molar mass of carbon, which is approximately 12 g/mol.
Moles of Carbon = Mass of Coal / Molar mass of Carbon = 630 g / 12 g/mol = 52.5 mol
Since the heat released is -394 kJ per mole of carbon, we can find the total heat released by multiplying the heat per mole by the number of moles of carbon.
Heat (q) = (-394 kJ/mol) × (52.5 mol) = -20,655 kJ
So, when our lump of coal burns, it releases a wicked -20,655 kJ of heat! It's getting hot in here!
Now, let's move on to heating some water. How much water are we talking about? Let's find out!
To calculate the mass of water that can be heated, we can use the equation q = mcΔT, where q is the heat released, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The change in temperature is 100°C - 27°C = 73°C
The specific heat capacity of water is approximately 4.18 J/g°C (I know, I switched to J, bear with me here).
Now, we can rearrange the equation to solve for the mass of water.
m = q / (c * ΔT)
Now let's plug in the values!
m = -20,655 kJ / (4.18 J/g°C * 73°C)
After converting kJ to J and doing the math, we find:
m ≈ -20,655,000 / 305,14 ≈ -67.75 g
Oh no! It looks like there is an error in the calculation. I apologize for any confusion. It seems I made a mistake!
To find the heat (q) when a lump of coal is burned, we can use the equation:
q = m * ΔH
where q is the heat, m is the mass of the coal, and ΔH is the heat of combustion per unit mass of coal.
First, let's calculate the mass of the lump of coal using its density and dimensions:
Density of coal = 1.5 g/cm^3
Volume of coal = 8 cm * 7.5 cm * 7 cm = 420 cm^3
Mass = Density * Volume = 1.5 g/cm^3 * 420 cm^3
Now that we have the mass of the lump of coal, we can calculate the heat (q) using the given heat of combustion per unit mass of coal:
q = Mass * ΔH
= (1.5 g/cm^3 * 420 cm^3) * (-394 kJ / 1000 g)
Make sure to convert the heat of combustion from kJ to J before multiplying it by the mass. Since the question asks for the answer in kJ, we'll divide the final answer by 1000 to convert it back to kilojoules.
For the second question, to find the mass of water heated, we need to use the equation:
q = m * c * ΔT
where q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Initial temperature of water (T1) = 27°C
Final temperature of water (T2) = 100°C
The change in temperature can be calculated as:
ΔT = T2 - T1 = 100°C - 27°C
The specific heat capacity of water (c) is approximately 4.18 J/g°C.
Now, we can rearrange the equation to isolate the mass (m):
m = q / (c * ΔT)
Substitute the calculated value of q from the previous question into this equation, along with the given values of c and ΔT. The resulting mass will be in grams.
Remember to always double-check and verify your calculations to ensure accuracy.
394 kJ x (630/12) = ?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for mass H2O. Remember to use q in J not kJ.
Using 12.011 for atomic mass C I obtained 20,666,000 J.
1. Did you substitute -20665.3 kJ convert to J. If so that's the problem. That is a + number when you heat the water.
2. If you used a + number then count the number of significant figures. That's often a problem with these on-line databases.
Let me know please and we can try something else if this doesn't fix the problem.