Question: Using a Joule's calorimeter to investigate the efficiency of the conversion of electrical energy to heat energy.

Equipment:
Joule's calorimeter
balance
thermometer (0-100degrees celsius)
power supply (0-12V)
ammeter (0-5A)
rheostat
stopwatch
electrical leads and distilled water

We carried out the prac to find the efficiency of the calorimeter with a copper coil, But we had a slight hick-up, we kept getting over 100% efficiency instead of under 100%. This isnt possible because we're not turning heat energy into electrical energy. Here are our results after we did the prac:

Initial temperature = 23degrees
Final temperature = 44.5degrees
Temperature equilibrium = 21.5degrees
Mass of distilled water = 0.094kgs
Mass of calorimeter = 0.122kgs
Current (I) = 2.2A
Potential difference (V) = 3.8V
Time = 880seconds
Specific heat capacity of water = 4180
Specific heat capacity of copper = 390
Relevant equations
P = IVt
Q=mc (delta) t

My attempts at the solution

P = 3.8*2.2*880
P = 7356.8W or 7356.8Js

Q(copper+water) = (0.0943*4180*21.5) + (0.122*390*21.5)
Q(copper+water) = 9497.7

Efficiency:

9497.7/7356.8 x 100% = 129.1% efficiency. How is this so? shouldn't it be lower.

Does anyone know what could be the problem/s. How to fix this issue? we tried 5 times and cept getting over 100% efficiency, Is there anything we can change to correct the efficiency?

Thank you

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The effects of sonoluminescence due to cavitation in electricity driven rotary systems such as Joe Grigg's Hydrosonic pump also produce heat energy at over 100% efficiency.


http://www.youtube.com/watch?v=yh_-DUKQ4Uw


EM induction systems can be quite efficient at converting flux "eddies" into heat.