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Q: Fomula to calculate the cost of heating water ( Answered ,   1 Comment )
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 Subject: Fomula to calculate the cost of heating water Category: Science > Physics Asked by: nzsixthsense-ga List Price: \$5.00 Posted: 20 Jul 2005 22:46 PDT Expires: 19 Aug 2005 22:46 PDT Question ID: 546087
 ```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!```
 Subject: Re: Fomula to calculate the cost of heating water Answered By: redhoss-ga on 21 Jul 2005 06:48 PDT Rated:
 ```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```
 nzsixthsense-ga rated this answer: `Awesome - thank you. That's exactly the answer I was looking for.`

 Subject: Re: Fomula to calculate the cost of heating water From: greendesigner-ga on 28 Nov 2005 19:07 PST
 ```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.```