Recent content by rthrbe

it is the time vs temperature graph

Can you please read the threads too, i have already shown my solution but I haven't figure out yet what really might be the other cause why my value is little bit higher.

I used the logger pro, and the gradient was 0.3808. I've used the formula E= cm delta T ofcourse,

i mean what could be the factors that affect my experiment, i mean why did i get 4, 33 instead of 4,18 there should be an explanation to that...

the specific heat capacity of water is 4, 18 kJ, that is the theoretical value, but on our experiment we got 4, 33 kJ, it is pretty close, and apparently there should be some kind of errors like loss of energy, something like that. i am not sure that's why i am asking for help

Yeah i mean i could have some errors in my experiment, like the heat energy transferred (not sure w the term) not only to the water but also to the kettle. the electric kettle is made of plastic... is there any factors or errors?

I had an experiment to find the specific heat capacity of water. Materials are electric kettle, logger pro, 1 kg water, and set up the time to 240 seconds. The experiment value i got is 4.33 kJ/(kg.K) which is closed to the waters specific heat capacity 4,18 kJ/(kg.K). I wonder what could be...