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At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula c = α + β T + δ T −2 , where α = 349 J/kg · K, β = 0.107 J/kg · K 2 , and δ = 4.58 × 10 5 J · kg · K . How much heat is needed to raise the temperature of a 2.00-kg piece of copper from 20 ° C to 250 ° C ?

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In a calorimeter of negligible heat capacity, 200 g of steam at 150 ° C and 100 g of ice at −40 ° 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 ° C is not enough to condense the steam, but it is more than enough to lower the steam’s temperature by 50 ° C , so the final state will consist of steam and liquid water in equilibrium, and the final temperature is 100 ° C ; 9.5 g of steam condenses, so the final state contains 49.5 g of steam and 40.5 g of liquid water.

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An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white ( e = 0 ) except for a 5 cm × 8 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 W/m ° C , and its inner surface is at a temperature of 20.0 ° 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.

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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. d L / d T = k T / ρ L ; b. L = 2 k T t / ρ L f ; c. yes

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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 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?

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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.

  1. In terms of Earth’s radius, the constant σ , and the unknown temperature T s of the surface, what is the power of the infrared radiation from the surface?
  2. What is the power of Earth’s radiation absorbed by the atmosphere?
  3. In terms of the unknown temperature T e of the atmosphere, what is the power radiated from the atmosphere?
  4. Write an equation that says the power of the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits.
  5. Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits.
  6. Solve your two equations for the unknown temperature of Earth.
    For steps that make this model less crude, see for example the lectures by Paul O’Gorman.

a. σ ( π R 2 ) T s 4 ; b. e σ π R 2 T s 4 ; c. 2 e σ π R 2 T e 4 ; d. T s 4 = 2 T e 4 ; e. e σ T s 4 + 1 4 ( 1 A ) S = σ T s 4 ; f. 288 K

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Questions & Answers

two point charges +30c and +10c are separated by a distance of 80cm,compute the electric intensity and force on a +5×10^-6c charge place midway between the charges
Tijani Reply
what is the difference between temperature and heat
Ishom Reply
Heat is the condition or quality of being hot While Temperature is ameasure of cold or heat, often measurable with a thermometer
Temperature is the one of heat indicators of materials that can be measured with thermometers, and Heat is the quantity of calor content in material that can be measured with calorimetry.
2. A brass rod of length 50cm and diameter 3mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°c( degree Celsius) if the original length are 40°c(degree Celsius) is there at thermal stress developed at the junction? The end of the rod are free to expand (coefficient of linear expansion of brass = 2.0×10^-5, steel=1.2×10^-5k^1)
A charge insulator can be discharged by passing it just above a flame. Explain.
Mudassar Reply
of the three vectors in the equation F=qv×b which pairs are always at right angles?
what is an ideal gas?
Justine Reply
What is meant by zero Kelvin ?
Why does water cool when put in the pot ?
when we pour the water in a vessel(pot) the hot body(water) loses its heat to the surrounding in order to maintain thermal equilibrium.Thus,water cools.
when we drop water in the pot, the pot body loses heat to surrounded in order to maintain thermal equilibrium thus,water cool.
types of thermometer?
yemisi Reply
thermometer, Radiation thermometer and vapour pressure thermometer.liquid thermometer use thermometric liquid like mercury ,alcohol etc.
liqid thermometer ,gas thermometer, resitance thermometer,thermo electric thermometer , radiation thermometer andvapour pressure thermometer
calculate the quantity of heat required to rise the temperature of 1gmail of ice _10 to 110
Dargu Reply
A 40cm tall glass is filled with water to a depth of 30cm. A.what is the gauge pressure at the bottom of the glass? B.what is the absolute pressure at the bottom of the glass?
Abdulaziz Reply
A glass bottle full of mercury has mass 50g when heated through 35degree, 2.43g of mercury was expelled. Calculate the mass of the mercury remaining in the bottle
Anjorin Reply
Two electric point charges Q=2micro coulomb are fixed in space a distance 2.0cm apart. calculate the electric potential at the point p located a distance d/2 above the central point between two charges
Abdul Reply
what is wave
Ahmed Reply
A wave is a periodic disturbance which travel with a finite velocity and remains unchanged in type as it travels.
What's a wave motion?
What is charge bodies
Oje Reply
which have free elections
Show that if a vector is gradient of a scaler function then its line around a closed path is zero
Charge bodies are those which have free electons
the melting point of gold is 1064degree cencius and is boiling point is 2660 degree cenciu
Ilyas Reply
is Thomas's young experiment interference experiment or diffraction experiment or both
Ilyas Reply
An aqueous solution is prepared by diluting 3.30 mL acetone (d = 0.789 g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?
A 4.0kg mess kit sliding on a fractionless surface explodes into two 2.0 kg parts.3.0 m/s due to north and 0.5 m/s 30 degree north of east. what is the speed of the mess kit
Practice Key Terms 9

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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