Heat loss via convection

Suppose I have a glass container (trough) and I want to heat the water inside. I put it on a hot plate. Now, even though the hot plate is at $110\,^{\circ}C$, the liquid only reaches $50\,^{\circ}C$. I want to know why.

The heat loss to the surroundings is convection or radiation. If I assume only radiation, the numbers don't add up. Can you tell me how to calculate the heat lost through convection (by air) based on area, temperature etc (An estimate would be good). Also evaporation might play a role too.

Please guide me on how to understand this.

• There is no easy way to calculate convective heat loss. Sep 7, 2015 at 6:03
• 10 - 100 W / m^2 K Sep 7, 2015 at 7:12
• You question is not clear. Is the plate kept at a constant temperature of 110 °C? What is the initial temperature of the liquid? May 9, 2016 at 13:02

Heat transfer texts clearly indicate that empirical equations are involved, due to the fact that turbulence is often encountered. You may have to do a search for your particular geometry to find the equation that estimates your convective heat loss.

Disclaimer: This answer doesn't answer your question about convection (which may explain the down vote). Nevertheless, I hope it can help you solve your problem about heat loss. If the method gets you close to your actual heat loss, you may be able to side-step convection. But first, try to compute the heat loss by estimating a convection rate in the range provided by Frisbee's comment.

Convection is difficult to model. It depends on convection cells which can be caused by random non-uniformities in the water and air.

An alternative is to calculate heat loss through (1) evaporation at the water/air interface, and (2) radiation at the glass/air interface.

You'll need to compare the rate of loss with the rate of heat conduction through the bottom of the glass container, in order to determine if heat coming through the bottom is being radiated and vaporized away rapidly enough to explain the temperature difference between hot plate and water.

In other words, by how much does the rate of heat going out exceed the rate of heat coming in?

First, you need to know the rate of Heat conduction from hot plate through the glass to the water:

Q / t = k * A * (T1 - T2) / d

Q is quantity of heat units

t is unit time

k is thermal conductivity of the glass trough (Here is a link to a table of thermal conductivity for various materials, including glass.)

A is area of the bottom of the trough

T1 is temperature of the hot plate surface

T2 is temperature inside the glass at bottom of the trough

d is thickness of the glass

Now, you can compute the rate of heat loss:

(1) Radiation: Calculate thermal radiation from the sides of the trough to the ambient air by using the Stefan-Boltzmann law.

(2) Evaporation: As heat is extracted from water in the trough by the phase change from liquid to vapor, you lose heat by evaporation, . Here is a link to a calculator for heat loss through vaporization.

Add the rate of heat loss from vaporization and radiation. Subtract that from the rate of thermal conductivity through the glass bottom, and see if it explains the temperature difference.

Convection carries heat through the water, and once it leaves the surface, through the air. But you may be able to explain much of the heat loss without quantifying convection.