### USB coffee-warmers: waste of money

The line of gimmicky things-that-plug-into-USB gadgets includes: USB-powered mug warmers. These are pointless, and here’s why: USB doesn’t provide enough power to make any appreciable difference.

USB ports are rated at up to 500mA at 5V: that’s a whole 2½W of warming power.

How much power does it take to keep a mug of tea warm? Well, let’s start with some assumptions. We’ll ignore the mug. We’ll assume tea’s pretty much water. We’ll guess that a mug holds about 250ml of tea; so a mug of tea has a mass of about 0.25 kg. Tea’s brewed at 100°C, drinkable at 60°C. And let’s say a mug of tea cools from just-brewed to drinkable in 10 minutes.

Now, the specific heat capacity of liquid water is 4186 J/kgK. So in cooling from 100°C to 60°C, the mug of tea has lost:

It’s lost that energy over 10 minutes, which is 600 seconds; the tea is losing heat at an average power output of:

That’s way, way more than the feeble 2.5W the warmer is putting into the tea.

However, averaging over time assumes that objects cool linearly. Not so: Newton’s Law of Cooling states that the rate of heat loss, and hence the rate of temperature change, is proportional to the temperature difference between the body and its surroundings:

We can solve this for the proportionality constant K by integrating over time. Let’s assume the surroundings are at 20°C:

This means that at 60°C, the rate of cooling is:

The heat loss per second is:

Newton helps a bit—it’s easier to keep a mug of tea at a lower temperature than at a higher temperature—but still: it’s going to take a lot more than this product’s 2.5W of power to do so.

In fact, let’s use Newton to calculate how warm 2.5W of heating is going to keep the mug of tea:

2 degrees above ambient. USB coffee warmers: cute gimmick, useless product. If you want to keep your tea warm, a covered mug would make a much bigger difference.

My back-of-the-envelope calculations are borne out by Amazon’s reviews of electric coffee-warmers. Take for example this Rival unit, rated at 22W, which reviewers still claim is underpowered: “barely warms the cup, let alone the contents”.

USB ports are rated at up to 500mA at 5V: that’s a whole 2½W of warming power.

How much power does it take to keep a mug of tea warm? Well, let’s start with some assumptions. We’ll ignore the mug. We’ll assume tea’s pretty much water. We’ll guess that a mug holds about 250ml of tea; so a mug of tea has a mass of about 0.25 kg. Tea’s brewed at 100°C, drinkable at 60°C. And let’s say a mug of tea cools from just-brewed to drinkable in 10 minutes.

Now, the specific heat capacity of liquid water is 4186 J/kgK. So in cooling from 100°C to 60°C, the mug of tea has lost:

ΔQ = mcΔT = 0.25 x 4186 x (100-60) = 4186 J of energy.

It’s lost that energy over 10 minutes, which is 600 seconds; the tea is losing heat at an average power output of:

W = ΔQ/Δt = 4186/600 = 67W.

That’s way, way more than the feeble 2.5W the warmer is putting into the tea.

However, averaging over time assumes that objects cool linearly. Not so: Newton’s Law of Cooling states that the rate of heat loss, and hence the rate of temperature change, is proportional to the temperature difference between the body and its surroundings:

dT/dt = -K(T-T

_{s})We can solve this for the proportionality constant K by integrating over time. Let’s assume the surroundings are at 20°C:

K = (1/t

_{1})ln(T_{0}-T_{s}/T_{1}-T_{s}) = (1/600)ln(100-20/60-20) = 0.00116.This means that at 60°C, the rate of cooling is:

dT/dt = -K(T-T

_{s}) = -0.00116 x (60-20) = 0.046 °C/s.The heat loss per second is:

ΔQ = mcΔT = 0.25 x 4186 x 0.046 = 48 J/s = 48W.

Newton helps a bit—it’s easier to keep a mug of tea at a lower temperature than at a higher temperature—but still: it’s going to take a lot more than this product’s 2.5W of power to do so.

In fact, let’s use Newton to calculate how warm 2.5W of heating is going to keep the mug of tea:

dT/dt = -K(T-T

⇒ dQ/dt = -mcK(T-T

⇒ T-Ts = -1/mcK dQ/dt = 1 / (0.25 x 4186 x 0.00116) x 2.5 = 2.07°C.

_{s}) and ΔQ = mcΔT⇒ dQ/dt = -mcK(T-T

_{s})⇒ T-Ts = -1/mcK dQ/dt = 1 / (0.25 x 4186 x 0.00116) x 2.5 = 2.07°C.

2 degrees above ambient. USB coffee warmers: cute gimmick, useless product. If you want to keep your tea warm, a covered mug would make a much bigger difference.

My back-of-the-envelope calculations are borne out by Amazon’s reviews of electric coffee-warmers. Take for example this Rival unit, rated at 22W, which reviewers still claim is underpowered: “barely warms the cup, let alone the contents”.

Second paragraph says "that’s a whole 2½W of warming power." Everywhere else you refer to "the feeble 0.25W the warmer is putting into the tea".

I think the 2.5W is correct (P=IV)...

It's still feeble though...