I need a forumla for calculating the cost of heating water in a hot
water tank based on the following variables:

* intial temperature of water supplied to the tank (assume 8 degrees Celcius)
* final temperature of the water in the hot water tank (assume 60 degrees Celcius)
* The volume of new water in the tank being heated to the final
temperature (assume 150 litres)
* The kilowatt rating of the heater elment in the hot water tank (no
idea yet - assume 0.3kW as an example)
* The cost per kilowatt (assume 15 cents per kilowatt). 

So basically I need to know what the total energy required in
kilowatts to heat a set volume of water from an initial temperature to
a final temperature which I can then multiply by the electricy unit
cost.

Ignore heat loss from the hot water tank - it's well insulated and I'm
after a general idea of cost rather than anything that needs to be too
acurate.

Please answer with a formula so that I can put it into a spreadsheet
and change the variables to get the correct answer.

In case you're intersted in why I'm asking this - I have 2 room mates
who are taking baths every single day and causing the electricity bill
to skyrocket.  I want to give them a real indication of how much
taking a bath costs so they can choose to pay extra for the additional
electricty cost.

Thank you!

Hello nzsixthsense, your question is a simple problem in
thermodynamics. I found a very good website that explains better than
I could myself:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/houseenergy.html

The energy necessary to heat water is determined by the specific heat of water:

cwater = 1 calorie/gm °C = 4186 J/kg°C = 1 BTU/lb °F
A typical U.S. residential water heater will be taken as one which has
a capacity of 40 U.S. gallons (= 320 pounds, 145 kg, 151 liters). The
typical heating range will be taken to be from 60 °F to 140 °F (15.6
°C to 60 °C). The energy required to heat the water can be determined
from the specific heat relationship.

Q = cm?T
The energy required to heat one tank of water over the specified range is then

(1 BTU/lb °F)(320 lb)(140°F - 60 °F) = 25,600 BTU

or

(4186 J/kg°C)(145 kg)(60 °C - 15.6 °C) = 26.9 million Joules 

Since a kilowatt-hour is 3.6 million Joules, this energy amounts to
about 7.5 KW of electricity. Taking an electric energy cost of
9.5¢/KW, it would cost about 71¢ to heat one tank of hot water with an
electric hot water heater assuming all the electric energy went into
heating the water.

For your parameters (except we will use their 151 liter number) the
answer would be:

(4186 J/kg°C)(145 kg)(60 °C - 8 °C) = 31.6 million Joules

31.6 million Joules / 3.6 million Joules = 8.8 KW

8.8 KW X .15 = $1.32 

The rating of the element in your water heater has nothing to do with
the cost. It only has to do with how fast the water is heater
(recovery time).

So, your bathers need to cough up a couple of bucks for every long
shower they take.

Hope this helps you out, Redhoss

Awesome - thank you.  That's exactly the answer I was looking for.

Nice explanation. I would just like to make one correction in how you
labeled your units of energy, as it is a common point of confusion.
Where you write kilowatts (KW) it should read kilowatt-hours (kWhr).
Kilowatts are units of power, while electricity is sold in units of
energy, kilowatt-hours. The numerical answer doesn't change.