Thermodynamics, energy, work, heat problem

[edsuave]

a system consisting of 73.2g of liquid water at 298k is heated using an immersion heater at a constant pressure of 1.00bar. if a current of 2.25A passes through the 10 ohm resistor for 125s, what is the final temp of the water?

im having trouble finding temp?
i found power and work using P = IR^2 and work = P/t
but I am stuck, i initially thought i should use PV = nRT but i have too many unknowns
please help asap



Moderator's Note: Multiple Threads Merged

 

[Redbelly98]

a system consisting of 73.2g of liquid water at 298k is heated using an immersion heater at a constant pressure of 1.00bar. if a current of 2.25A passes through the 10 ohm resistor for 125s, what is the final temp of the water?

im having trouble finding temp?
i found power and work using P = IR^2 and work = P/t

You should re-check that equation for work. It's not quite right.

The key is that the work is entirely converted into heat energy. So you'll need an equation that relates heat energy to temperature change.

... i initially thought i should use PV = nRT ...

Nope. That equation is for an ideal gas. The water is a liquid here.

 

[edsuave]

i know Heat = mass * Cp * change in temperature but how does the power relationship tie into this?

 

[mukundpa]

What is specific heat of a substance? Do you know the definition?

 

[edsuave]

Q=nC(Tf-Ti) i believe C is the specific heat

 

[mukundpa]

Rather for water (liquid) we can write Q = mC (Tf - Ti)

You have calculated Q, See for specific heat of water, it is a constant and calculate !

 

[Redbelly98]

i know Heat = mass * Cp * change in temperature

Good, that's right. Keep that one in mind.

but how does the power relationship tie into this?

There's another useful equation relating energy, power and time. See my previous comment about re-checking an equation you wrote in post #1.

 

[Borek]

You start with liquid water at 298 K - what will happen to it when it is heated?

 

[kingkool]

The formula you have Q = mC (Tf - Ti) only works if the heat capacity is constant throughout the range of temperatures. It might be close enough approximation in this case, if that's what your professor told you then use that equation. But in a general snse, you have to integrate from Tf to Ti...

 

[Redbelly98]

The formula you have Q = mC (Tf - Ti) only works if the heat capacity is constant throughout the range of temperatures.

We usually consider C to be a constant in introductory physics problems.

(And it is constant to within about 0.1% in this case.)