# Phys Chem Heating water with electric current

## Homework Statement

86.9g of liquid water at 304K is heated by 1.75A passing through 24.7Ω for 104s what is the final temp

## Homework Equations

its been a while since i did physics so bear with me if i get something mixed up here
q=mCs,pΔT q=ItΔψ ψ=pot dif so isnt that just V from v=IR?

also i believe the q=ItΔψ results in units as AVs if A=Cs-1 then we would get CV which is = to 1J???? long shot but maybe im correct

## The Attempt at a Solution

assuming the above is true i want to say that Tf=[(I2tR)/(Cs,pm)]+Ti

plugging in a bunch of numbers and if those units worked out how i think i get 325.65K

Does this seem legit?

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gneill
Mentor

## Homework Statement

86.9g of liquid water at 304K is heated by 1.75A passing through 24.7Ω for 104s what is the final temp

## Homework Equations

its been a while since i did physics so bear with me if i get something mixed up here
q=mCs,pΔT q=ItΔψ ψ=pot dif so isnt that just V from v=IR?
Looks good. It's handy to remember a few relationships for electric power when resistors are involved: P= VI, P = V2/R, and P = I2R.

also i believe the q=ItΔψ results in units as AVs if A=Cs-1 then we would get CV which is = to 1J???? long shot but maybe im correct
Sure. But since you're not given the potential difference (Voltage) across the resistor but your are given the current, you can use the current-version of the power expression. Then

$q = I^2R\:Δt~~~~~~~~~~~$where Δt is the elapsed time that the current flows.

## The Attempt at a Solution

assuming the above is true i want to say that Tf=[(I2tR)/(Cs,pm)]+Ti

plugging in a bunch of numbers and if those units worked out how i think i get 325.65K

Does this seem legit?
Sure. Not only that, but your results looks good