Thread: physics: heat transfer problem, copper sauce pan and boiling water

1. physics: heat transfer problem, copper sauce pan and boiling water

A copper bottomed sauce pan contained 0.800L of boiling water boils dry in 12.0 min. Assuming that all the heat is transferred through the flat copper bottom, which has a diameter of 15.0cm and a thickness of 3.70mm, calculate the temperature difference between the inside and outside of the copper bottom while some water is still in the pan.

I have no idea how I'm supposed to solve this. I have been calculating whatever I can out of the data, though which I will provide here, but be cautious because my calculations could be wrong.

Volume of Copper Pan bottom = (0.075^2)(pi)(0.0037) = 6.538 * 10^-5 cubic meters

Latent heat of evaporation of water = 2257 kJ/kg provided by engineertoolbox

mass of initial water= 0.800 kg

c_copper = 0.39 kJ/(kg*K) provided by engineertoolbox

power = (2257*0.800)/(12*60) = 2.507777 kW

energy required to evaporate 0.800kg of water = 0.800 * 2257 = 1805KJ

2. Re: physics: heat transfer problem, copper sauce pan and boiling water

Hmm sounds like you need to use Conduction Equation:
P = kA(Th-Tc)/l = dE/dt
As well as Calorimetry Equation:
Q = Lm = LpV (p is density of water)
dQ/dt = Lp(dV/dt) = dE/dt

kA(Th-Tc)/l = Lp(dV/dt)
Th-Tc = Lpl/(kA)*(dV/dt)

For this, you will need to know the k value for copper. dV/dt will just be the volume of liquid, .800L divided by how much time it completed boiled off, 12mins.