Would adding water after heating decrease the overall heat, when compared to adding the water before heating for the same period of time?

I was rather cold last night, and warmed up a cup of water to drink in the microwave. I put it in for 60 seconds, but it came out boiling hot, so I put a bit of cold water in with the hot water, and it was warm for me to drink.

So then it came to me - what if I had put that extra water in before I put it in the microwave? If I still heated it for 60 seconds, would the extra surface area of the extra water cause it to heat faster, or would the speed remain the same, due to the averaging of heat?

Thanks.

• The real question is - People warm their cups of water in a microwave? People drink warm water? As for your question, I believe the 60 seconds in the microwave after you add more water will make the temperature change lower. $Q=mc\Delta T$. Same time period means the heat, $Q$, is the same (need confirmation about that), $c$ is the specific heat (constant) and $m$ is what you're changing. We see that, under these conditions, $\Delta T\; \alpha\; \frac1{m}$, so more mass means less temperature change. – Shahar May 11 '14 at 1:25

It's safe to approximate that all of the microwave's energy is deposited in the water. To some extent you will also heat up the mug and the walls of the microwave, but we will neglect that in favor of a bigger effect.

If the water reached 100 ºC in the microwave before the heating cycle ended, the heat after that was "wasted" by creating water vapor.

Suppose what you want is water at 80 ºC. You pour some room-temperature water into a cup and microwave it for 60 s. After 40 s its temperature reaches 100 ºC; after this its temperature stops changing and vapor begins forming. At the end of the 60 s you remove the boiling water and add more room-temperature water. Because the heat of vaporization for water is quite large compared to the heat capacity for the liquid, very little of the boiling water actually left the cup as steam in the microwave. Essentially the final third of the microwave's heat is wasted.

Now suppose instead you add enough water that it would take 75 seconds to boil, but only put it in the microwave for 60 s. Voilà: all of the microwave's heat usefully raises the temperature of the water.

A microwave works by dielectric heating, where an oscillating high frequency electric field causes polar molecules to align in the field. Since the field alternates, they will rotate continuously thereby dispersing energy in the process macroscopically observed as temperature.

The quantity of heat $Q$ will not remain constant for different objects being heated$^{1}$. The cup prior to heating will therefore receive the same $Q$ per unit mass; thus, it will have a higher temperature in the end.

$^{1}$ Not entirely sure about this, and I don't have a thermometer to test this experimentally.

It will come out cooler if you add the extra water at the end. It could be a lot-if you literally mean it came out boiling hot at the end of 60 seconds, you could have used the last 15 seconds to boil water. If you had added the extra water at the beginning, it would come out at 100C. If you add it at the end the mix will be cooler. The extra heat in the first case went into boiling water.

Even if you never boil the water, the temperature at a given time will be hotter for the smaller quantity. You will then evaporate more water, losing heat in the process. Adding the water at the end will still result in a cooler final temperature, but probably only a small amount.

It would most likely be less warm if you add more water. It is because heat is measured in calories or temp. 1 calorie is how much energy that is required to heat up 1cm3 water up one degree Celsius. So theoretically if I added more mass or water it would take more energy to heat up because 25 cm3 of water would less energy to heat up because it has less mass than 50 cm3 of water. Heat doesn't change so time is the variable.

Let us say for example you heated up half a cup of chocolate milk (25 cm3) for one minute and it would take x amount of heat compared to a full cup of chocolate milk (50 cm3) which might take 1.5x or 2x the time or heat that half a cup of chocolate milk took to reach same temp. It simply requires more energy and calories to heat them to the same temp...

Let's go back of the envelope here: I'm going to assume that your "original" volume of water is 6 oz (I think that's fairly standard for a cup of coffee/tea), but you add an "extra" 2 oz of cold water. (I'll refer to the amounts of these waters by the quoted titles throughout the answer to keep the language precise). Whether or not this water will be added before/after microwaving will be left to be seen momentarily. Also, let's assume a 1000 Watt microwave, to keep the math easy, and because it splits the difference nicely between the common 900 Watt and 1100 Watt models. Finally, I'll use your 60 seconds as the time interval, also I can assure you my water never gets hot that fast in the microwave.

If your 1000 Watt microwave is left on for one minute, the total work done by the microwave will be $W=P\Delta t$. That means that over 60 seconds, your microwave does 60000 Joules of work. In both this part of the answer, as well as the other, I will assume that all of this work is transferred into the water as heat. $$Q=mC\Delta T$$ where $Q$ is the heat, $m$ is the mass of the water, $C=4.18 J/g$ per degree Celcius is the specific heat of water, and $\Delta T$ is the temperature change we're looking for. $$60000J=(237g)(4.18 J/g)\Delta T$$ When we crunch numbers, we get $$\Delta T=60.6$$ degrees Celcius. If our water started at a room temperature of 22 degrees Celcius, that gives us a final temperature of $T_f=82.6$ degrees Celcius. Ok, so perhaps we were too optimistic to think that all of the microwave's work heated the water, but at least we've got a reasonable temperature to compare against the next step.
For this step, let's heat only our 6 ounces of original water in the microwave. Because we only have 6 ounces of water in the microwave, the mass of the water will now only be 177 grams, but otherwise our specific heat calculation will be identical to find $\Delta t$. $$Q=mC\Delta T$$ again, which will look like $$60000J=(177g)(4.18J/g)\Delta T$$ where $$\Delta T=81.09$$ Oh shoot! Looks like when we consider the starting temperature of room temperature water, our water will begin to experience a phase change in the microwave. However, the heat of fusion is quite large, compared to the small amount of heat which would have raised our water above 100 degrees Celcius, so I will just assume that our water temperature caps at $T_f=100$ degrees Celcius without a significant loss of mass to steam. Now the fun part. We will add 2 ounces, or 59 grams of room temperature ($T=22$) water to our microwaved water. When the hot, original water and the room temperature, extra water are mixed, they will exchange heat with each other such that $$Q_{original} +Q_{extra}=0$$ so that $$m_{original}C\Delta T_{original} =-m_{extra}C\Delta T_{extra}$$ but $C$, the specific heat of water divides out so $$(177g)(100-T_f)=-(59g)(22-T_f)$$ and $$17700-(177g)T_f=-1298+(59g)T_f$$ $$18998=(236g)T_f$$ where we get that $T_f=80.5$ degrees Celcius. This is slightly lower than our answer for "Adding Water First"