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539 calories per gram for heat of vaporization plus 1 cal/gram/degree C

100 degrees C - 80 degrees C = 20 degrees C

(539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal

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Wiki User

12y ago
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Wiki User

13y ago

Use this formula, then convert.

q( in Joules ) = mass * specific heat * change in temp.

q = (5000 grams)(4.180 J/gC)(100 C - 20 C)

= 1672000 Joules

Now, 4.184 Joules = 1 calorie, so....

1672000 Joules (1 calorie/4.184 Joules)

= 399617.59 calories ( you do significant figures )

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Wiki User

13y ago

Energy = mass x specific heat capacity x change in temperature

The specific heat capacity(the energy required to raise 1 kg of a substance by 1 degree Celsius) of water is around 4200 Joules per kilogram per degree Celsius. The change in temperature is 100- 80 = 20 degrees Celsius.

Energy = 50 x 4200 x 20 = 4.2 x 105 Joules

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Wiki User

7y ago

Approx 1132 kiloJoules.

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Wiki User

7y ago

117,180,000

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Anonymous

Lvl 1
3y ago

117,180,000 j

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Q: How much heat energy is needed to convert 50 kg of water at 80 degrees Celsius to steam at 100 degrees Celsius?
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