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.