Expert: Josh - 2/23/2007

Question
Question
a perfectly insulated container contains 70 g of water at 40 oc. A
small heater is immersed in the water and is connected to a 12 V
battery. An electric current of 13 A flows through the heating
element. The specific heat of water is water is 4.2 x 10+ to the
power of 3) J kg-1  oC(celcius) -1 and the latent heat of
vaporization is 2.6 x 10+ to the power of 6) J kg-1.
(a)i need to know how to calculate the time taken for the water to reach boiling temperature
(b) calculate the time taken to vaporize all the water.
But i need to know how they are calculated in all the steps in the process

Answer
Hi Cara,

This question is about using electrical energy to heat a substance. Assuming no loss of energy during heating, the specific heat of water tells us that 4.2 kJ is required to raise the temperature of 1 kg of liquid H2O (water) by 1 degree. Furthermore, when water reaches boiling point, chemistry teaches us that to transform from the liquid state to a gaseous state (forming steam), the water molecules need a massive 2.6 MJ (Mega Joules) of energy to "get free". This is what is referred to as latent heat.

DATA
Mass of water:
 m=70 [gram]
Temperature of water:
 T=40 [Celsius]
Voltage source:
 V=12 [Volt]
Current:
 I=13 [Ampere]
Specific heat of water:
 K(water)=4200 [J/kg/C]
Latent heat of vaporization of water:
 L(water->steam)=2600000 [J/kg]
a)

Step 1: The heating device generates energy. Electrical energy is consumed by the water. Power is given by P=VI (product of voltage and current flowing through the heating element). Thus, P=156.

Step 2: Calculate energy needed to reach 100 degrees Celsius.
We need to raise the temperature of water from 40 to 100 (by 60 degrees). Using the specific heat of water,
the required energy is:
"Mass of Water" x "Temperature Gradient" x "Specific Heat" or
E(required)
=m*(T_final-T_initial)*K(water)
=0.070 [kg] * (100-40) [C] * 4200 [J/kg/C]

I'll let you simplify this yourself. Note: be careful with the units, e.g., when you convert from gram to kilogram, and bear in mind that 1 MJ = 1,000 kJ = 1,000,000 J.

Finally, time taken to boil water is the total energy required, divided by the rate which this energy is delivered by the heating element, viz., the electric power.

t(required)=E(required)/P

b) The ideas are very similar. Time taken equals L(water)*m divided by P. How do you figure this out? Hint: Check the dimensions of the variables in the formula.

Latent heat L(water) is measured in [J/kg]
mass of water "m" is measured in [kg]
When the two multiplies, the dimensional units [kg]/[kg] cancels out. leaving you with energy measured in [J]

Energy [J] = Power [J/s] * Time [s]
Rearranging, Time [s] = Energy [J] / Power [J/s]

Makes sense?