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

Answer 1

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

Answer 2

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 )

Answer 3

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

Answer 4

To convert 50 kg of water at 80 degrees Celsius to steam at 100 degrees Celsius, you need to calculate the heat energy needed to raise the temperature of water from 80 to 100 degrees Celsius (specific heat capacity of water) and then the heat energy needed for water to steam phase change (latent heat of vaporization of water). The total heat energy required can be calculated using the formula: Q = mcΔT + mL, where Q is the heat energy, m is the mass, c is the specific heat capacity, ΔT is the temperature change, and L is the latent heat of vaporization.

Answer 5

Approx 1132 kiloJoules.

Answer 6

117,180,000

Answer 7

117,180,000 j