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Heating water with amps ohms and time 
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#1
Sep2413, 12:40 AM

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so if i have 89.6g water at 304K and a constant p=1.00bar and i heat it by running 1.75A through 24.7 for 105 seconds what will the final temp be?
im thinking i can take q=mC(T_{f}T_{i}) and q=IT and I=R/V to say that T_{f}= (Rt/vCm)+T_{i} i cant rember that much about physics and this stuff isnt in my book, yet its on my study list...First off will this work. it appears that it would. Secondly what units would one use to do this the best i can figure the unit work would be something like (Ω*s)/(v*K^{1}g^{1}*g) but this is one of those funny things where i dont know what that corresponds to 


#2
Sep2413, 12:49 AM

P: 17

ya i screwed that up bad I=V/R and I dont have any value for V so no help...what on earth do you do to something like this?



#3
Sep2413, 02:02 AM

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'running 1.75A through 24.7 for 105 seconds'
24.7 what? You need to review how power is calculated knowing electric current: http://en.wikipedia.org/wiki/Electric_power Hint: 1 volt times 1 amp = 1 watt of power. 


#4
Sep2413, 02:34 AM

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Heating water with amps ohms and time



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