Heat transfer to solution per litre

[rwooduk]

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.

 

[mjc123]

The units of specific heat capacity are J g^-1 K^-1. So the equation is dimensionally correct. But if the saucepan is being "heated at constant temperature", dT/dt = 0. So what do you mean here?
How to proceed depends on the assumptions you are making. Are you assuming no heat is lost from the water?
Are you assuming the heat capacity of the pan is negligible compared with that of the water? Are you assuming a constant power, in which case the rate of temperature change will decrease as the mass of water increases, or are you trying to get a constant dT/dt, in which case the power must increase as the mass increases?
If you want the power per unit volume, you must divide by the volume.
P/V = dT/dt*cp*M/V = dT/dt*cp*ρ where ρ is the density of water in g/L.

 

[rwooduk]

Thank you very much for the reply. My mistake on the units of specific heat, also I meant the heat is constant from the stove and heats the water.

The calorimetric (experimental) measurement is based purely on the change in temperature of the water, so heat capacity of the pan, heat loss from the water, change in the mass of water (it doesn't reach boiling point) etc does not factor in.

So for 400 mL of water with a measured calorimetric power of 31.92 W, for W/L I must divide by 0.4L?

EDIT I think I have it now.

Thanks again.