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At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula $c=\alpha +\beta T+\delta {T}^{\mathrm{-2}},$ where $\alpha =349\phantom{\rule{0.2em}{0ex}}\text{J/kg}\xb7\text{K,}$ $\beta =0.107\phantom{\rule{0.2em}{0ex}}\text{J/kg}\xb7{\text{K}}^{2},$ and $\delta =4.58\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\text{J}\xb7\text{kg}\xb7\text{K}\text{.}$ How much heat is needed to raise the temperature of a 2.00-kg piece of copper from $20\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ to $250\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ ?
In a calorimeter of negligible heat capacity, 200 g of steam at $150\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ and 100 g of ice at $\mathrm{-40}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present?
The amount of heat to melt the ice and raise it to $100\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ is not enough to condense the steam, but it is more than enough to lower the steam’s temperature by $50\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , so the final state will consist of steam and liquid water in equilibrium, and the final temperature is $100\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ ; 9.5 g of steam condenses, so the final state contains 49.5 g of steam and 40.5 g of liquid water.
An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white $\left(e=0\right)$ except for a $5\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}8\phantom{\rule{0.2em}{0ex}}\text{cm}$ patch in the form of the astronaut’s national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity $k=0.0600\phantom{\rule{0.2em}{0ex}}\text{W/m}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , and its inner surface is at a temperature of $20.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hard to solve in closed form, so you can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.
The goal in this problem is to find the growth of an ice layer as a function of time. Call the thickness of the ice layer L . (a) Derive an equation for dL / dt in terms of L , the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at $t=0$ , you have $L=0.$ If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get dL / dt multiplied by a (very simple) function of L on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on t . (c) Will the water eventually freeze to the bottom of the flask?
a. $dL\text{/}dT=kT\text{/}\rho L$ ; b. $L=\sqrt{2kTt\text{/}\rho {L}_{\text{f}}}$ ; c. yes
As the very first rudiment of climatology, estimate the temperature of Earth. Assume it is a perfect sphere and its temperature is uniform. Ignore the greenhouse effect. Thermal radiation from the Sun has an intensity (the “solar constant” S ) of about $1370\phantom{\rule{0.2em}{0ex}}{\text{W/m}}^{2}$ at the radius of Earth’s orbit. (a) Assuming the Sun’s rays are parallel, what area must S be multiplied by to get the total radiation intercepted by Earth? It will be easiest to answer in terms of Earth’s radius, R . (b) Assume that Earth reflects about 30% of the solar energy it intercepts. In other words, Earth has an albedo with a value of $A=0.3$ . In terms of S , A , and R , what is the rate at which Earth absorbs energy from the Sun? (c) Find the temperature at which Earth radiates energy at the same rate. Assume that at the infrared wavelengths where it radiates, the emissivity e is 1. Does your result show that the greenhouse effect is important? (d) How does your answer depend on the the area of Earth?
Let’s stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity $e=0.77$ (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun’s radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun’s radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere’s transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere’s radius is not significantly different from Earth’s, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth’s area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth’s surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth’s surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere.
a. $\sigma \left(\pi {R}^{2}\right){T}_{\text{s}}{}^{4}$ ; b. $e\sigma \pi {R}^{2}{T}_{\text{s}}{}^{4}$ ; c. $2e\sigma \pi {R}^{2}{T}_{\text{e}}{}^{4}$ ; d. ${T}_{s}^{4}=2{T}_{e}^{4}$ ; e. $e\sigma {T}_{\text{s}}^{4}+\frac{1}{4}(1-A)S=\sigma {T}_{\text{s}}^{4}$ ; f. 288 K
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