Find the final temp (Physical chemistry)

In summary, the final temperature of the water can be found by using the equation for heat capacity and substituting the value for electrical work instead of heat, due to the fact that the change in internal energy of the closed system is equal to the sum of heat and work.
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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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First of all, great job on solving the problem correctly! Your understanding of thermodynamics seems to be on the right track.

To answer your question, yes, you are correct in your reasoning that delta(U) = q + w = 0 in this case. This means that the change in internal energy of the system is equal to the heat transferred into the system (q) plus the work done by the system (w), and in this case, the net change is zero.

In this problem, the electrical work done by the immersion heater is the only form of work being done on the system, and since it is a closed system (no mass exchange), the heat added to the system (q) and the electrical work done (w) are equal in magnitude, but opposite in sign. This is why you were able to use the equation for heat capacity (q/delta(T)) and substitute the value for w instead of q.

I hope this helps clarify your confusion. Keep up the good work!
 

1. What is the formula for finding the final temperature in physical chemistry?

In order to find the final temperature, you can use the formula: Tf = (m1c1T1 + m2c2T2) / (m1c1 + m2c2), where Tf is the final temperature, m1 and m2 are the masses of the substances, c1 and c2 are their respective specific heats, and T1 and T2 are their initial temperatures.

2. How do you determine the specific heat of a substance?

The specific heat of a substance can be determined by measuring the amount of heat required to raise the temperature of a unit mass of the substance by 1 degree Celsius. This value is usually given in units of J/g°C.

3. Can the final temperature be negative?

Yes, the final temperature can be negative if the initial temperatures of the substances are different and the mass and specific heat of one substance is significantly greater than the other.

4. How does the mass of a substance affect the final temperature?

The larger the mass of a substance, the more heat it can absorb or release, which can affect the final temperature. This is accounted for in the formula for finding the final temperature.

5. What are some real-life applications of finding the final temperature in physical chemistry?

Finding the final temperature is important in many real-life scenarios, such as in cooking and baking, determining the temperature change in a chemical reaction, and understanding the heat exchange in a thermodynamic system.

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