The change in temperatures over the course of the year. Click image to enlarge. Images from 1987 via NOAA's GLOBE Earth System Poster.
The continents heat up faster than the oceans, and they cool down faster too. You can see this quite clearly in the animation above: notice how cold North America gets in the winter compared to the North Atlantic. It’s why London has an average January low temperature of 2˚C while Winnepeg’s is closer to -20˚C, even though they’re at almost the same latitude. There are a few reasons for the land-ocean cooling differences, and they all have to do with how heat is absorbed and transported.
(1) Specific Heat Capacity. Water has a higher heat capacity than land. So it takes more heat to raise the temperature of one gram of water by one degree than it does to raise the temperature of land. 1 calorie of solar energy (any type of energy really) will warm one gram of water by 1 degree Celcius, while the same calorie would raise the temperature of a gram of granite by more than 5 degrees C. The Engineering Toolbox has specific heat capacities of common materials.
(2) Transparency. The heat absorbed by the ocean is spread out over a greater volume because the oceans are transparent (to some degree). Since light can penetrate the surface of the water the heat from the sun is dispersed over a greater depth.
(3) Evaporation. The oceans loose a lot of heat from evaporation. In the evaporative heat loss experiment, While there is some evaporation from wet soils and transpiration by plants, the land does not have anywhere near as much available moisture to cool it down.
(4) Currents. Not only do the oceans absorb heat over a greater depth, but they can also move that energy around with their currents. The solar energy absorbed at the equator gets transported towards the poles, while the colder polar water gets transported the other way. Currents help average out ocean temperatures.
The inner core of the Earth is made of solid metal, mostly iron. The outer core is also made of metal, but it’s liquid. Since it formed from the solar nebula, our planet has been cooling down, and the outer core has been freezing onto the inner core. Somewhat counter-intuitively, the freezing process is a phase change that releases energy – after all, if you think about it, it takes energy to melt ice.
The energy released from the freezing core is transported upward through the Earth’s mantle by convection currents, much like the way water (or jam) circulates in a boiling pot. These circulating currents are powerful enough to move the tectonic plates that make up the crust of the earth, making them responsible for the shape and locations of the mountain ranges and ocean basins on the Earth’s surface, as well as the earthquakes and volcanics that occur at plate boundaries.
Conceptual drawing of assumed convection cells in the mantle. (via The Dynamic Earth from the USGS).
Eventually, the entire inside of the earth will solidify, the latent heat of fusion will stop being released, and tectonics at the surface will slow to a stop.
The topic came up when we were talking about the what heats the Earth. Although most of the energy at the surface comes from solar radiation, students often think first of the heat from volcanoes.
Note: An interesting study recently published showed that although the core outer core is mostly melting, in some places it’s freezing at the same time. Unsurprising given the convective circulation in the mantle.
Model of convection in the Earth's mantle. Notice that some areas on the mantle are hotter, creating hot plumes, and some are cooler (image from Wikipedia).
Note 2: Convection in the liquid outer core is what’s responsible for the Earth’s magnetic field, and explains why the magnetic polarity (north-south) switches occasionally. We’ll revisit this when we talk about electricity and dynamos.
This simple experiment was devised to estimate just how much heat is lost from a teacup due to evaporation as compared to the other types of heat loss (conduction and convection).
Experimental setup for measuring evaporative heat loss.
The idea is that if we can measure the mass of water that evaporates over a short period of time, we can calculate the evaporative heat loss because we know that the amount of heat it takes to evaporate one gram of water (its latent heat of evaporation) is 540 cal/g. So we’ll take some hot, almost boiling, water and weigh and take its temperature as it cools down.
Materials
Apparatus.
It requires:
A thermometer (Celcius up to 100 degrees)
A styrofoam cup (because it’s light)
A digital scale (to take quick measurements to tenths of a gram)
A 100 ml graduated cylinder (optional)
A beaker (100 ml) or cup that can go in the microwave
water
Procedure
Our scale has a capacity of about 120 g so we need to make sure that the combined weight of our apparatus that will go on the scale is less. The plan is to have the styrofoam cup, with a thermometer and some water on the scale. Since we can be somewhat flexible with how much water is in the cup we’ll first weigh the cup and thermometer.
(my measurement, not necessarily yours)
Mass of styrofoam cup and thermometer = 29.6 g
So it should be safe to use 70 g of water, which is approximately equal to 70 ml since the density of water is 1 g/ml.
1. Measure the 70 ml of water in the graduated cylinder and put it into the beaker (or microwavable cup). The exact volume is not crucial here since we’ll be using the scale to measure the mass of water more precisely.
2. Microwave the water for about 40 seconds. Again you do not have to be too precise here, you just want the water to be close to boiling. The length of time you need to microwave the water will depend on the strength of your microwave. 40 seconds raised the temperature of my 70 ml of water from 22˚C to 82˚C. If you like you can calculate the heat absorbed by the water, and the effective power of the microwave from these numbers, but it is not necessary for this experiment.
3. Quickly place the hot water into the styrofoam cup with the thermometer on the scale and measure the mass and the temperature of the water.
4. Measure the mass and temperature of the water every 2 minutes for the next 10 minutes.
Calculations
1. Every time you took a measurement, the temperature and the mass should have dropped. The change in mass is due to evaporation. Every time one gram evaporated, 540 calories are lost. Calculate the amount of heat lost due to evaporation at every time measurement.
Hint: Evaporative Heat Loss = mass evaporated × latent heat of evaporation
QE = mE LE
2. Now that you know how much heat was lost, you can figure out how much of the temperature drop was caused by evaporation. Since the specific heat capacity of water is 1 cal/g/˚C, each calorie lost due to evaporation should have reduced the temperature of one gram of the water by one degree Celsius.
Hint: Evaporative Temperature Change = Evaporative Heat Loss × mass of water in container × specific heat capacity of water
∆TE = QE / (m Cw)
You should also the temperature drop caused by evaporation as a percentage of the total temperature drop. Hopefully, your result is less than 100%.
There are quite a number of things that might come up in discussion here, for example: just how large are the potential for measurement errors; and are the results comparable to an actual teacup.
My trial of this experiment indicated that about 69% of the heat loss was due to evaporation. It should be possible to also calculate the amount of heat loss from conduction through the walls of the cup; the thermal conductivity of styrofoam is 0.033 W/mK (via the Engineering Toolbox). The radiative heat loss can be estimated using Stefan’s Law, which can be used to account for all the different methods of heat loss.
Finally, there is no control described in this experiment. A useful thing to try would be to use a styrofoam cup with a lid.
Additional Notes
When my students tried this experiment they use a small (50ml) beaker and 25g of water. Their evaporative heat loss was only 44% of the total, probably due to the smaller volume of water, which as a larger surface-area to volume ratio, and the thinner, more conductive glass walls of the beaker.
