Recent content by kendreaditya

Yes, I understand that because increasing the SA is inversely proportional to the equilibration time. This being the work case scenario, and having a heating time of 12 min is not terrible, but it may be an area of concern. Background/purpose: The reason I need to know this because I am...

Thank You! I am having trouble understanding the mean of "after 4 time constants", does this mean I have to have 4-time constants? or a time constant of 4? So what would be a good way to calculate the heat transfer coefficient or calculate an upper bound, as this equilibration time ranges...

So it would take around 9 mins to equilibrate?

I got the value from https://www.engineeringtoolbox.com/overall-heat-transfer-coefficients-d_284.html Also, I am still getting a different number now.

What I am thinking is to use Newtons law of cooling. where t (time constant) is C/Ah (3.3595) C = heat capacity (385 J/kgC) A = heat transfer surface area (28.65 cm^2) h = heat transfer coefficient (400 W/(m2 K) copper to water) deltaT(t) = 0.01C deltaT(0) = 75C I got 385/(400 * .2865) =...

That is correct, I estimated a little less because when I use it in the environment, the surface area will decrease. That 28.65 cm^2 is the actual SA.

I have a copper cylinder that is at 0C, how much time will be needed for it to equilibrate with 10kg of water at 75C? Copper cylinder: Mass: 105.7 g Specific heat: 0.385 J/gC Initial Temp: 0C Final Temp: 75C ish SA: 25 cm^2 Water: Mass: 10000g Specific heat: 4.184 J/gC Initial Temp: 75C...