Phys Chem Heating water with electric current

[speny83]

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=mC_s,pΔT q=ItΔψ ψ=pot dif so isn't 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 I am correct

The Attempt at a Solution

assuming the above is true i want to say that T_f=[(I²tR)/(C_s,pm)]+T_i

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

Does this seem legit?

 

[gneill]

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=mC_s,pΔT q=ItΔψ ψ=pot dif so isn't 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 I am 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 T_f=[(I²tR)/(C_s,pm)]+T_i

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