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, Sirsh.