This type of question is pretty common in introductory engineering thermodynamics and chemical energetics courses. They can be quite challenging if you aren't used to deriving the information that has been omitted in the question.
The first step is to make a couple of assumptions:
- The heaters power output is constant
- No heat is lost to the environment (The process is 'Adiabatic')
Alright, now for the fun part. Using the first law of thermodynamics, the conservation of energy, we know that all of the energy coming from the heater is transferred to the water, and we know that it takes 1 calorie to raise 1g of water by 1°C. Ok, so what good is that? We are still missing the rate of energy coming from the heater and also the mass of the water.
Well, as it turns out, we can deal with this problem independent of the mass.
We can now say that the change in specific internal energy (Δu) of the water is proportional to the change in it's temperature (ΔT). This forms the 1st law expression:
Δu = specific heat x ΔT
Δu = 1cal/g°C x (100 - 50)°C
Δu = 50 cal/g
This means that the heater must be supplying 10 calories/g every minute. Now that we at a point where the water is boiling, all of the energy goes into releasing the water molecules from the liquid state instead of raising the temperature. Conveniently enough, they have provided you with a value that tells you how much energy this requires:
- Heat of vaporization - 540 cal/g
Using these two bits of information, we see that if we have 10 calories coming in each minute and we need 540 calories then the process is going to take 540/10 mins or 54 mins. Having said that, because of the lack of mass, we must remember that this is a specific value that means it will take 54mins/g of water.
Well, I hope that clears things up a bit for you. They may have intended something different but that's the answer that the question leads to.
Good luck on your entrance exam!
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Topic: Specific heat question (Read 3322 times)