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?

Question

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?

Answer 1

First, convert 100 lbs ice to grams. 100 lbs x (454 g/lb) = ?grams. You may want to use 453.6 g/lb or 453.59 g/lb..

Second, determine q needed to heat the ice & water.
q1 = heat needed to melt ice.
q1 = mass ice x heat fusion.

q2 = heat needed to raise liquid water from zero C to 100 C.
q2 = mass water x specific heat H2O x (Tfinal-Tinitial)

q3 = heat needed to convert water at 100 to steam at 100.
q3 = mass H2O x heat vaporization.
Total q = q1 + q2 + q3.

Then,
2.54E4 J/g x #g coal = q to heat H2O.
Substitute and solve for #g coal.
Post your work if you get stuck.

Answer 2

Does an answer around 5.36563 kg sound about right? After adding q1+q2+q3 I got 136287 kj so I converted that to J and then found the #g coal to be 5.36563 kg.

Answer 3

To find the quantity of coal required, we need to calculate the energy needed to convert 100 pounds of ice to steam.

Step 1: Convert the weight of ice from pounds to grams.
1 pound = 453.592 grams
So, 100 pounds = 100 * 453.592 = 45,359.2 grams.

Step 2: Calculate the heat energy required to convert the given quantity of ice to steam.
The energy required can be calculated using the formula:

Energy = Mass * Specific Heat Capacity * Temperature Change

For the given scenario, the specific heat capacity of ice is 2.09 J/g°C, and the temperature change is from 0.0 degrees Celsius to 100.0 degrees Celsius.

Energy = 45,359.2 * 2.09 * (100 - 0) = 9,488,053.28 J

Step 3: Convert the energy from joules to grams.
The given heat of combustion of bituminous coal is 2.5 x 10^4 J/g.

Quantity of coal required = Energy required / Heat of combustion

Quantity of coal required = 9,488,053.28 / (2.5 x 10^4) grams

Quantity of coal required = 379.522 grams

Therefore, approximately 379.522 grams of bituminous coal is required to produce the energy needed to convert 100 pounds of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius.