- #1
rwooduk
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- Homework Statement
- This is a question I made up to understand more about the calorimetric power. The equation below is given by Koda et. al (2003).
A saucepan of water (400 mL) on a stove is been heated at constant temperature. The calorimetric power can be written:
##Power(W)= (\frac{dT}{dt}) c_{p}M##
Where (dT/dt) is the temperature rise per second (0.019 °C/s), ##c_{p}## is the heat capacity of water (4.2 J/g), M is the mass of water (400 g).
Many papers give the calorimetric power in W/L, which I think is just the same as the above calculation.
So, the question, what is the power here in W/L?
- Relevant Equations
- ##Power(W)= (\frac{dT}{dt}) c_{p}M##
Problem Statement: This is a question I made up to understand more about the calorimetric power. The equation below is given by Koda et. al (2003).
A saucepan of water (400 mL) on a stove is been heated at constant temperature. The calorimetric power can be written:##Power(W)= (\frac{dT}{dt}) c_{p}M##Where (dT/dt) is the temperature rise per second (0.019 °C/s), ##c_{p}## is the heat capacity of water (4.2 J/g), M is the mass of water (400 g).
Many papers give the calorimetric power in W/L, which I think is just the same as the above calculation.
So, the question, what is the power here in W/L?
Relevant Equations: ##Power(W)= (\frac{dT}{dt}) c_{p}M##
Firstly, I am confused by the equation, by dimensional analysis we have:
##W= (\frac{°C}{s})(\frac{J}{g}) g = (\frac{J}{s})(°C) = W (°C) ##
So the temperature part seem to be remaining.
Now the solution:
Power(W) = (0.019)(4.2 J/g)(400 g) = 31.92 W
We can see from the equation that for an increase in volume of water (M) the rate of change of temperature would decrease. As expected, more water, less of a temperature increase. Therefore wouldn't the rate of energy transfer to solution be the same for 400 mL and 1L?
I see two ways to go:
a) Divide both sides of the equation by 1L to give W/L. But this would mean that the value for power(W) is the same as the power in W/L i.e. 31.92 W/L. Which would be correct assuming the same energy transfer to 400 mL as 1 L.
b) If we take the power(W) as the power in 400 mL of solution, then we simply multiply W by (1/0.4) to get the power in 1L of solution. But power is not the power in solution, it is the energy transfer to solution.
I think I am making this more complicated than it needs, or I have made a mistake somewhere, so would really appreciate any guidance.
A saucepan of water (400 mL) on a stove is been heated at constant temperature. The calorimetric power can be written:##Power(W)= (\frac{dT}{dt}) c_{p}M##Where (dT/dt) is the temperature rise per second (0.019 °C/s), ##c_{p}## is the heat capacity of water (4.2 J/g), M is the mass of water (400 g).
Many papers give the calorimetric power in W/L, which I think is just the same as the above calculation.
So, the question, what is the power here in W/L?
Relevant Equations: ##Power(W)= (\frac{dT}{dt}) c_{p}M##
Firstly, I am confused by the equation, by dimensional analysis we have:
##W= (\frac{°C}{s})(\frac{J}{g}) g = (\frac{J}{s})(°C) = W (°C) ##
So the temperature part seem to be remaining.
Now the solution:
Power(W) = (0.019)(4.2 J/g)(400 g) = 31.92 W
We can see from the equation that for an increase in volume of water (M) the rate of change of temperature would decrease. As expected, more water, less of a temperature increase. Therefore wouldn't the rate of energy transfer to solution be the same for 400 mL and 1L?
I see two ways to go:
a) Divide both sides of the equation by 1L to give W/L. But this would mean that the value for power(W) is the same as the power in W/L i.e. 31.92 W/L. Which would be correct assuming the same energy transfer to 400 mL as 1 L.
b) If we take the power(W) as the power in 400 mL of solution, then we simply multiply W by (1/0.4) to get the power in 1L of solution. But power is not the power in solution, it is the energy transfer to solution.
I think I am making this more complicated than it needs, or I have made a mistake somewhere, so would really appreciate any guidance.
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