CATAGORY: Thermodynamics
QUESTION: The heating element of an electric hot plate, 26.5-ohms
resistance, is connected across a 115-V circuit. (a)How long will it take
to heat 750 g of water at 20.0 degrees celcius to its boiling point, 100
degrees celcius, assuming no loss of heat to the surroundings? (b)What
power is dissipated in this heating element?

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The specific heat capacity of water is 4186 J/kg C.  the energy required
to raise .750 kg of water from  20.0C to 100.0C is

(1) E = mass(temperature change)specific heat
    
      = .750kg(100.0C -20.0C)4186J/kg C

      = 251160 J ~ 2.51 x 10^5 J
      
The energy imput into the water per unit time is

(2) P = V^2/R

where V is the voltage and R the resistance.

thus

(3) P = (115-V )^2/26.5-ohms = 499 J/s

(note: If the source is a.c. then the power is reduced by a factor of 2.)

the time is then

(4) time = Energy/power

         = E / P
         
         = 2.51 x 10^5 J / 499 J/s
         
         = 5.030 x 10^2 s 
         
         ~ 8.38 min
      
         
 ***********************
 *                     *
 * time = 8min 23s     *
 *                     *
 * P = 499 W           *
 *                     *
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