Heating water with amps ohms and time

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To determine the final temperature of 89.6g of water at 304K after applying a current of 1.75A through a resistance of 24.7 ohms for 105 seconds, the discussion highlights the need to calculate the heat added to the water using the formula q = mC(Tf - Ti). The power can be calculated using the electric power formula, where power (in watts) is the product of voltage and current (P = IV). The current and resistance can be used to find voltage (V = IR). The participants emphasize the importance of understanding unit conversions and the relationships between electrical power, heat transfer, and temperature change. The conversation underscores the need for clarity on what "24.7" refers to, as it is crucial for calculating the voltage and subsequently the heat energy transferred to the water.
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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(Tf-Ti) and q=IT and I=R/V

to say that Tf= (Rt/vCm)+Ti

i can't rember that much about physics and this stuff isn't 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-1g-1*g) but this is one of those funny things where i don't know what that corresponds to
 
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ya i screwed that up bad I=V/R and I don't have any value for V so no help...what on Earth do you do to something like this?
 
'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.
 
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