Energy balance problems

Note: I don't know how to get html to do superscripts, so I use the symbol ^ before exponents. Thus, for example, J/(m^2 s^1) is read joules per square meter per second.

1) Consider a leaf that is absorbing 370 W/m^2 (read, watts per meter squared) of solar radiation. If it is losing 180 W/m^2 by the evaporative cooling effects of transpiration, how much heat must it exchange (gain or lose?) by convection and conduction in order to reach a stable temperature?

The basic issue here is that in order to maintain a stable body temperature, the leaf needs to lose heat as fast as it is absorbing it. Thus it needs to lose a total of 370 W/m^2. The leaf will then lose (370 - 180) = 190W/m^2, by a combination of convection, conduction, and thermal (infrared) radiation.

2) You're lying on a beach on a sunny day, absorbing 600 joules per square meter of skin per second [written 600 J/(m^2 s^1)]. Assume that your convective and conductive heat exchange are in equilibrium with your surroundings, and that you are sooooo relaxed that your metabolic heat production is negligible. To evaporate 1 gram of water uses 2259 J. Assuming that you have an exposed skin surface area of 1m^2 (1 square meter, a reasonable approximation), how much sweat would you have to evaporate in an hour to maintain a stable body temperature?

Note that in this problem I give the input from solar energy in J/ (m^2 s^1), but in the first problem I used W/m^2. Because one watt is one joule per second (1 W = 1 J/s^1), these measures [J/(m^2 s^1) and W/m^2] are the same thing. Now, how do we solve this problem? First, recall the energy balance terms. In this problem, convection and conduction are set to zero. Radiation input (from the sun) is 600 J/(m^2 s^1) or, since we specified that you have a skin surface area of 1 m^2, your heat input is 600 J/s^1 (equal to 600 Watts, i.e. six 100W lightbulbs). So to maintain a stable temperature, you're going to have to dispose of that heat just as fast.

Each gram of water evaporated takes 2259 joules. So 1 g/s^1 (gram per second) of evaporation would have a cooling effect of 2259J/s^1. But you only need 600 J/s^1 of cooling or 1 g times (600/2259) per second, or 0.266 g/s^1. That is 0.266 x 60 = 15.9 g/min^1 (grams per minute), and 15.9 x 60 = 956 g/hr^1. How much is that? It is the water in the sweat that evaporates. One gram of water has a mass of 1 mL (milliliter), so that would be would be just under 1 liter. The standard size soft drink can is 355 ml, so you'd need to evaporate nearly three of those per hour to stay cool.

3) Continuing from the preceding question - assume you have a mass of 65 kg. Assume that your mass is entirely water (a very rough approximation). Each gram of water contains 4.2 joules of heat energy for each degree Celsius of temperature. If in the preceding example, you sweat 500 ml of water in one hour (1 ml water = 1 gram), by how much would your body temperature change? Would it increase or decrease?

Evaporating 500 g water per hour, at 2259 joules per gram consumes 314 joules per second. That leaves a heat uptake of 286 joules per second. 65 kg = 65000 grams.

286 joules per second / 65000 grams = 0.004404 joules per gram per second, multiplied by 3600 s/hr is 15.9 joules per gram per hour. Since the heat of 1 g of water is 4.2 joules/¾C^1, that would result in heating of 3.8¾C/hr^1.