The heat of combustion of bituminous coal is 2.5 x 10^4 J/g. What quantity of coal is required to produce the energy required to convert 100. Pounds of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius? Is the answer 5.36 about right. I'm not sure if I did it correctly
Then why not post your work and let us check it?
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100lbs x453.6g/lbs = 45360grams
45360 grams x 1mol/18.02 g= 2517.2 moles
2517.2mol x 6.01kj/mol = 15128.4kj
45360 grams x 4.18j/g C x 100C x 1/1000 = 18960.5
45360g x 1 mol/ 18.02g x 40.6 kj/ mol = 102198kj
All added equals 136287 kj
2.54E4 J /1g x #g = 136287 kj
#g= 5.36 kj
I found your answer to my response last night and I expect you are ok. I used 2257 J for heat vaporization and 334 for heat fusion and 4.184 for specific heat H2O and came up with 5460 g,
Your 40.6 kJ/mol is about 2253 J/g and I used 2257.
I used 4.184 and you used 4.18.
Your 6.01 kJ/mol is 333.5 J/g and i used 334 for that. My answer came out 5460 g coal. Your procedure is good and my assumption is that the different numbers we used is responsible for the slight difference in g coal.
Thank you very much!!!
To solve this problem, you need to determine the energy required to convert 100 pounds of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius.
Here are the steps to calculate it:
Step 1: Convert pounds to grams.
To convert pounds to grams, multiply the value by 453.592 (1 pound = 453.592 grams). Therefore, 100 pounds is equal to 453.592 x 100 = 45,359.2 grams.
Step 2: Calculate the heat energy required to convert the ice to steam.
The specific heat capacity of ice is 2.09 J/g °C, and the heat of vaporization of water is 2.257 x 10^3 J/g. Therefore, to calculate the energy required to convert the ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius, you need to consider two processes:
- Heating the ice from 0.0°C to 100.0°C:
Energy = mass (in grams) x specific heat capacity x ΔT
Energy = 45,359.2 g x 2.09 J/g °C x (100.0°C - 0.0°C)
- Vaporizing the water at 100.0°C:
Energy = mass (in grams) x heat of vaporization
Energy = 45,359.2 g x 2.257 x 10^3 J/g
Step 3: Calculate the quantity of coal required.
The heat of combustion of bituminous coal is given as 2.5 x 10^4 J/g. To find the quantity of coal required, divide the total energy required (from Step 2) by the heat of combustion of coal:
Quantity of coal = Total energy required ÷ Heat of combustion of coal = (Energy from Step 2) ÷ (2.5 x 10^4 J/g)
Calculating the above equation will give you the quantity of coal required in grams.
Now, let's calculate the answer:
Step 1: Convert pounds to grams:
100 pounds = 45,359.2 grams (as calculated earlier)
Step 2: Calculate the energy required:
- Heating the ice from 0.0°C to 100.0°C:
Energy = 45,359.2 g x 2.09 J/g °C x (100.0°C - 0.0°C) = 9,460,894.08 J
- Vaporizing the water at 100.0°C:
Energy = 45,359.2 g x 2.257 x 10^3 J/g = 102,326,742.4 J
Total energy required = 9,460,894.08 J + 102,326,742.4 J = 111,787,636.48 J
Step 3: Calculate the quantity of coal required:
Quantity of coal = 111,787,636.48 J ÷ (2.5 x 10^4 J/g) = 4,471.505458 grams ≈ 4,472 grams (rounded to the nearest whole number)
Therefore, the quantity of coal required to produce the energy to convert 100 pounds of ice to steam is approximately 4,472 grams.
Based on your answer of 5.36, it seems that you have made an error in the calculations. Please double-check the steps and units used in your calculations to identify the source of the error.
You type in the work like the rest of us.
Give me a second and I'll work the problem and see if 5.36 is about right.