Expert: Josh - 4/30/2004

Question
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.

I need to know:
1.calculte the energy required for the water to reach boiling
temperature (100 degrees) ?

(2)calculate the energy required for the water to reach boiling
temperature?

(3)calculate the time taken for the water to reach boiling
temperature?

(4)calculate the energy to vaporize all the water?

(5)calculate the time taken to vaporize all the water?

all answers must assume that all the electrical energy is transferred
to heat the water, and the answers to two significant figures would
be appropiate.

any algebra equations to go along with the answers would be accepted
most gratefully, thankyou...

[Difficulty]
puuting it all into the right equations..

[Thoughts]
question 1, the power supplied, 13 A (AMPS)? X 70g of water ? = 910 J
kg -1 oC (CELCIUS)-1?

I CANT WRITE THE ALGEBRA IN, I HAVE NOT GOT THE SYMBOLS ON MY
PC>>>>>>>>


Answer
Hi Andrew,

I really appreciate that you explained your difficulties.
Okay, the specific heat tells us the amount of energy needed to raise the temperature of a substance by 1 degree celsius per unit mass.

In this problem, we convert (transform) electrical energy into chemical energy.
The electrical power (meaning dissipated energy per unit time [J/sec]) is given by P=VI, where V is the electric potential (also called "voltage") and I stands for electric current, measured in [Ampere].
So, the battery supplies 12*13=156 J/s.

Write "k_water(aq)" for the specific heat of liquid water.
k_water(aq)=4.2x10^3 [J.K/kg]
Write "k_latent(water->steam)" for the latent heat of vaporization.
k_latent(water->steam)=2.6x10^6 [J/kg]

Let's measure temperature in Kevin [=celsius+273.15]
Temperature gradient, dT=T(target) - T(initial)=[373-(273+40)]=60 K

To bring a volume of water from 40 to 100 celsius, we need
E = {(m_water)*k_water(aq)*dT} / efficiency      [in Joules]

where volume of water is "v_water"=70 cm^3 or 70 ml,
density of water "rho_water" is assumed to be 1.0 g/cm^3,
mass of water is given by "m_water"=(v_water)*(rho_water)=0.070 [kg].
under the ideal assumption, 100% energy conversion efficiency means "efficiency=1" and there is no loss of energy in the process.

Ans 2: E = 0.070*4.2x10^3*60 = 1.7640x10^4 [Joules] (i.e., 1.8*10^4 to 2 sig.fig.)
Ans 3: Time taken to reach boiling pt. = E/(V*I) = 113.076...sec (you can round the answer yourself)
Ans 4: Need to supply a great deal of energy to overcome the inter-molecular force between the water molecules, to liberate them from one another.
E(aqueous-to-gaseous)= (m_water)*k_latent(water->steam) [Joules]
= 0.070 * 2.6x10^6 = 1.82x10^5 [Joules]
N.B. Same idea for sublimation in other problems (going from solid to gas).
Ans 5: same approach as part 3.

Note 1: Useful things to remember, if ever in doubt, check the dimensions of each parameter.
The units are there to help you.

eg., in part 2), the three quantities in the following equation,
E = (m_water)*k_water(aq)*dT
have the units, [kg], [J.K/kg], [K] respectively.
When they multiply together, common units cancel just like fractions. You're left with [J] in this case, which is a measure of energy.

Note 2: Some variations --
#If the electrical circuit in this problem consists of a resistor, R [Ohm], and a current I is passing through it.
The electrical power (energy/time) is given by P=(I^2)*R.

#If the electrical circuit connects a battery across a resistor, R,
then, it generates P=(V^2)/R [Joules/sec].

Cheers,
Josh