Heating water with amps ohms and time

[speny83]

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 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

 

[speny83]

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?

 

[SteamKing]

'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.

 

[Borek]

https://www.physicsforums.com/showthread.php?t=712277

 

[LudusRex]

.

Hello,

I would like to clarify that the equation you have used, q=IT, is the formula for calculating the heat generated by an electric current. However, in this case, we are looking at the heat required to raise the temperature of water, which is given by the equation q=mC(Tf-Ti) where q is the heat, m is the mass of the water, C is the specific heat capacity of water, Tf is the final temperature, and Ti is the initial temperature.

To find the final temperature, we can rearrange the equation as Tf= (q/mC) + Ti. Since we know the mass of water (89.6g), specific heat capacity of water (4.184 J/gK), and initial temperature (304K), we can calculate the heat required using the formula q=IT. We can also calculate the resistance (R) using the formula R=V/I, where V is the voltage (24.7V) and I is the current (1.75A).

Once we have the heat and resistance, we can plug them into the equation Tf= (Rt/vCm)+Ti to find the final temperature. The units for this equation would be in Kelvin (K) since we are dealing with temperature, and the other units will cancel out.

I hope this helps clarify your understanding. It is important to remember to always use the correct equations and units when solving scientific problems.