Question

The heat of combustion of bituminous coal is 2.50 × 104 J/g. What quantity of the coal is required to produce the energy to convert 106.9 pounds of ice at 0.00°C to steam at 100.°C? specific heat (ice) = 2.10 J/g°C specific heat (water) = 4.18 J/g°C heat of fusion = 333 J/g heat of vaporization = 2258 J/g

A)	5.84 kg
B)	0.646 kg
C)	0.811 kg
D)	4.38 kg
E)	1.46 kg

the answer is A I just want to know why? So show each formula please with the calcuations plugged in

Answer 1

106.9 pounds =48489.024 grams

first you have to melt the ice

(106.9 lbs)(453.6g/lb)(333J/g) = 1.61x10^7 J

Then you have to heat it to 100C

(100C)( 48489g)(4.18J/gC) = 2.03x10^7 J

Then boil the water

(48489g)(2258J/g) = 10.9x10^8 J
Now we add them all up and you get
14.54x10^7 J

The heat of combustion of bituminous coal is 2.50 × 104 J/g

To calculate quantity of the coal is required to produce the energy to convert 106.9 pounds of ice at 0.00°C to steam at 100.°C we use following conversation:

14.54x10^7 J / 2.50 × 104 J/g

=5816 g

= 5.84 kg