How many joules are required to heat 200.0 grams of water from 25 degrees C to 125 degrees C?

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There are a number of steps involved in heating water from 25 degrees C (298 degrees K) to 125 degrees C (398 degrees K). Water turns into steam (gas phase) at 100 degrees C (373 degrees K). The steam will then be heated from 100 degrees C to 125 degrees...

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There are a number of steps involved in heating water from 25 degrees C (298 degrees K) to 125 degrees C (398 degrees K). Water turns into steam (gas phase) at 100 degrees C (373 degrees K). The steam will then be heated from 100 degrees C to 125 degrees C. These are the steps involved in calculation:

1.) Heating water from 25 degrees C to 100 degrees C:

Energy needed = mass of water x specific heat capacity of water x temperature change = 200 g x 4.184 J/g/K x (373 - 298) K = 62,760 J.

(note that 75 degrees C increase in temperature is same as 75 K change).

2.) Evaporation of water or conversion to steam:

Energy needed = mass of water x latent heat of evaporation 

= 200 g x 2259 J/g = 451,800 J

3.) Heating steam from 100 degrees C to 125 degrees C:

energy needed = mass of steam x specific heat capacity of steam x temperature change 

= 200 g x 2.02 J/g/K x (398-373) K

= 10,100 J

Thus, the total energy needed = 62,760 + 451,800 + 10,100 

= 524,660 J or 524.66 kJ.

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