A Hot Cup of Coffee

Get Me a Cup of Coffee!

How much power does it take to heat up a cup of coffee?

Specific Heat of Water - (OK - coffee, but coffee's mostly water)

4.186 Joules/gram-Deg C

Specific Heat of the Coffee Cup - (Which weighs about half as much as the water....)

1.09 Joules/gram-Deg C

8 oz coffee weighs about 225 grams

The cup weighs about 100 grams

So - how much POWER (in Watts)  do I need to get my coffee and cup from room temperature (about 68 degrees which is 20 Deg C)

up to a piping hot, but drinkable 158 Deg F, which is 70 Deg Celsius.

in 1 minute?

Now - It turns out that 1 Joule = 1 watt second.

The change in temperature = 70 deg C - 20 Deg C = 50 Deg C

So for the coffee - it takes

4.186 Joules 1 Watt Second 1 Minute 50 Deg C x 225 grams

Watts = ----------------- X --------------------- X --------------- X ----------------------------- = 785 Watts

gram Deg C Joule 60 Seconds 1 Minute

For the cup - it takes

1.09 Joules 1 Watt Second 1 Minute 50 Deg C x 100 grams

Watts = ----------------- X --------------------- X --------------- X ----------------------------- = 91 Watts

gram Deg C Joule 60 Seconds 1 Minute

So it will take 785 + 91 = 876 Watts to get my coffee hot in one minute....

(This ignores, of course, the fact that my coffee cup is losing heat to the air and the counter it sits on all the time, so it will really take more power than this....)

Now it turns out that I can buy some Nickel Chrome wire to use for heating the coffee....

24 AWG NiCr 60

1.671Ω/Ft                    ~  200 ft roll:   $22.49

Link to buy a roll of 24 awg NiChrome wire

At 120 Volts, how much resistance to I need to produce 876 Watts?

Well... Power, P = I x E  - I know E is 120V, and I know the power is 876 Watts.

So now I re-write the equation so that:    I = P/E

But I want to know R -

I know R = E/I so I can re-write that equation  so that:  I = E/R

Now with SUBSTITUTION I get -

E/R = P/E.     Multiplying both sides by R gets me:    E = PxR/E

Multiplying now both sides by E gets me:  ExE= PxR

Dividing both sides by P gets me: ExE/P= R

So the resistance I need is:  120 volts x 120 volts / 876 Watts   = 16.4 ohms

So how many feet of this wire do I need?

16.4 ohms / 1.671 Ω/Ft = 9.81 Feet = 118 inches...

But WAIT! - That's the resistance at 20 degrees C -
What happens to the resistance of wire as it heats? It goes up.

So for a 50 degree C change in temperature, it turns out that the resistance will go up about 20%.

So at 70C I really have about 1.671*0.2 + 1.672 = 2.006 Ω/Ft

16.4 ohms / 2.006 Ohms/foot = 8.175 feet or 98.1 inches

Now if I wind this stuff in a 3/8” diameter helical coil pattern, allowing for one wire diameter width between the turns... I can get the length down to about 5 inches.

Wire Gage Dimensions

Online Helical Coil Calculator

So when I bend the coil in half, I can now fit it into the cup...

Now before I wire it  up to a plug, I’d better figure out if it’s too much current for the breaker!

Hmmm - I = E/R so the current will be about 120 volts / 16.4 Ohms  = 7.3 amps at 70C.

At room temperature, though, the current will be about 20% higher... or 7.3*1.2 = 8.8 amps.

So no worries there!  Ahhh!  A nice hot cup of coffee -

(* DANGER!!!  Don’t really do this - risk of electrocution!  this is just a thought problem! *)

That’s it for now