The heat of combustion of bituminous coal is 2.50 × 104 J/g. What quantity of th

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

A)

5.84 kg

B)

0.646 kg

C)

0.811 kg

D)

4.38 kg

E)

1.46 kg

Explanation / Answer

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

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