- #1

osker246

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## Homework Statement

A system of 57.g of liquid water at 298k is heated using an immersion heater at a contsant pressure of 1 bar. If a current of 1.50 A passes through the 10.0-ohm resistor for 150s. What is the final temperature of the water?

## Homework Equations

w=VIt

C=q/deta(t)

V=IR

C(h20)=75.3 (J/mol*k)

## The Attempt at a Solution

Ok so I solve the problem correctly but I am a little confused at one step. I will work through what I did to find the answer.

I found the potential using V=IR=(1.5)(10)=15 volts.

Since I now know the potential I found the electrical work, w=VIt=(15)(1.5)(150)=3375 J

Now where I am a little confused is to find my answer I essentially used the equation of heat capacity of water:

C=q/delta(T) --->T(final)=q/(C*n)+T(initial)=312k

But I didnt use heat (q) from the internal system, I used the value from electrical work (w). Thermodynamics is a fairly new topic for me so forgive me if this seems like a rediculous question. But why did I get the right answer if I used work instead of heat?

Is it because delta(U)= q + w = 0 and the system did work on the suroundings (so making it negative work), thus making q=w? I'd appreciate any help on this. Thanks.

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