a chemical engineer placed 1.520 g of a hydorcarbon in the bomb of a calorimeter. The bomb was immersed in 2.550 L of water and the sample was burned. The water temperature rose from 20.00C to 23.55C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat released per gram of hydrocarbon?
I'm unsure of my work but here it is:
Q(bomb) = 403J/K * 3.55K = 1430.65J
Q(H2O) = 2.550L * (1000g/L) * 3.55K * (4.184J/gK) = 37875.66J
Q(hydrocarbon) = -Q(bomb) -Q(H2O)
Q(hydrocarbon) = -(1430.65J) - (37875.66J) = -39306.31J
Final Answer:
1gram hydrocarbon = -39306.31J/1.520g = 25859.41J/g = 25900J/g
Does this seem right?
That looks ok to me although I didn't check the math.
yeah the answer is true.I'd checked it with my lecturer.
Hmm, let me calculate it in my own hilarious way!
So, a chemical engineer walks into a calorimeter with a hydrocarbon and says, "Watch me heat things up!" And boy, did things heat up!
First, let's calculate the heat absorbed by the bomb. We multiply the heat capacity of the calorimeter (403 J/K) by the temperature change (3.55 K). That gives us 1430.65 J.
Next, we check out the water. There's 2.550 L of it, and we all know water can be quite thirsty, so let's convert that to grams (1000 g/L). Then, we multiply it by the temperature change (3.55 K) and the specific heat capacity of water (4.184 J/gK). That gives us 37875.66 J.
Now for the hydrocarbon. The heat released by the hydrocarbon is the negative sum of the heat absorbed by the bomb and the heat absorbed by the water. We have -1430.65 J (because the hydrocarbon is a bit of an energy thief) and -37875.66 J (because the water just can't resist heat). Adding those together, we get -39306.31 J.
Finally, to find the heat released per gram of hydrocarbon, we divide -39306.31 J by the mass of the hydrocarbon (1.520 g). And after all that math and humor, we get approximately -25859.41 J/g, or let's just round it up to -25900 J/g.
So, it seems like you did a great job with your calculations! Keep up the good work, and remember, chemical reactions aren't just about the numbers, they're also about the chemistry puns!
Yes, your calculations are correct. The heat released by the hydrocarbon can be calculated by subtracting the heat absorbed by the bomb and the water from the total heat released.
Initial temperature of water (T1) = 20.00°C
Final temperature of water (T2) = 23.55°C
Change in temperature (ΔT) = T2 - T1 = 3.55°C
First, calculate the heat absorbed by the bomb:
Q(bomb) = heat capacity of the calorimeter (C) * ΔT
Q(bomb) = 403 J/K * 3.55 K = 1430.65 J
Next, calculate the heat absorbed by the water:
Q(H2O) = mass of water (m) * specific heat capacity of water (Cp) * ΔT
Since the volume is given, convert it to mass using the density of water:
mass of water (m) = volume of water (V) * density of water (ρ)
mass of water (m) = 2.550 L * 1000 g/L = 2550 g
Q(H2O) = 2550 g * 4.184 J/g°C * 3.55°C = 37875.66 J
Finally, calculate the heat released by the hydrocarbon:
Q(hydrocarbon) = -Q(bomb) - Q(H2O)
Q(hydrocarbon) = -(1430.65 J) - (37875.66 J) = -39306.31 J
To find the heat released per gram of hydrocarbon, divide the total heat released by the mass of the hydrocarbon:
Heat released per gram of hydrocarbon = Q(hydrocarbon) / mass of hydrocarbon
Heat released per gram of hydrocarbon = -39306.31 J / 1.520 g ≈ -25900 J/g
The answer is approximately -25900 J/g, or 25900 J/g (since the negative sign indicates the direction of heat transfer).
Your calculation seems mostly correct, but there is a slight error in the calculation of Q(H2O). Let's go through the calculation step by step to identify the error.
To find the heat released per gram of hydrocarbon, you need to calculate the total heat released and divide it by the mass of the hydrocarbon.
First, calculate the heat absorbed by the water (Q(H2O)):
Q(H2O) = mass of water × specific heat capacity of water × change in temperature
You correctly identified the mass of water as 2.550 L, but we need to convert it to grams since the specific heat capacity of water is given in J/g·K.
Water has a density of 1 g/mL, so 2.550 L is equivalent to 2550 g.
Now, calculate Q(H2O):
Q(H2O) = 2550 g × 4.184 J/g·K × (23.55 - 20.00) K
= 433,934.4 J
Next, calculate the heat absorbed by the bomb calorimeter (Q(bomb)):
Q(bomb) = heat capacity of the calorimeter × change in temperature
= 403 J/K × (23.55 - 20.00) K
= 1,416.65 J
Now, calculate the total heat released (Q(total)):
Q(total) = -(Q(H2O) + Q(bomb))
= -(433,934.4 J + 1,416.65 J)
= -435,351.05 J
Finally, divide the total heat released by the mass of the hydrocarbon (1.520 g):
Heat released per gram of hydrocarbon = -435,351.05 J / 1.520 g
= -286,245.82 J/g
It's important to note that the negative sign indicates that heat is being released by the hydrocarbon. Therefore, the correct answer is approximately -286,246 J/g.
So, there was a mistake in the calculation of Q(H2O) in your previous work, but now you have arrived at the correct answer.