What is the final temperature when the two become equal

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Reference no: EM131462450

Part -1:

1. How much heat, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?

2. An aluminum kettle weighs 1.05 kg.
(a) What is the heat capacity of the kettle?
(b) How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?
(c) How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)?

3. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal.

(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.

(b) The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.

4. When 1.0 g of fructose, C6H12O6(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?

5. How much heat is produced by combustion of 125 g of methanol under standard state conditions?

6. Calculate ΔH°298 for the process Zn(s) + S(s) + 2O2(g) ? ZnSO4 (s) from the following information:

Zn(s) + S(s) ?ZnS(s) ΔH°298 = -206.0 kJ
ZnS(s) + 2O2 (g) ? ZnSO4 (s) ΔH°298 = -776.8 kJ

Part -2:

1. Answer the following questions about a Blu-ray laser:

(a) The laser on a Blu-ray player has a wavelength of 405 nm. In what region of the electromagnetic spectrum is this radiation? What is its frequency?

(b) A Blu-ray laser has a power of 5 milliwatts (1 watt = 1 J s-1). How many photons of light are produced by the laser in 1 hour?

(c) The ideal resolution of a player using a laser (such as a Blu-ray player), which determines how close together data can be stored on a compact disk, is determined using the following formula: Resolution = 0.60(λ/NA), where λ is the wavelength of the laser and NA is the numerical aperture. Numerical aperture is a measure of the size of the spot of light on the disk; the larger the NA, the smaller the spot. In a typical Blu-ray system, NA = 0.95. If the 405-nm laser is used in a Blu-ray player, what is the closest that information can be stored on a Blu-ray disk?

(d) The data density of a Bluray disk using a 405-nm laser is 1.5 × 107 bits mm-2. Disks have an outside diameter of 120 mm and a hole of 15-mm diameter. How many data bits can be contained on the disk? If a Blu-ray disk can hold 9,400,000 pages of text, how many data bits are needed for a typed page? (Hint: Determine the area of the disk that is available to hold data. The area inside a circle is given by A = πr2, where the radius r is one-half of the diameter.

2. Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the Li2+ ion.

3. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a He+ ion moves from the orbit with n = 5 to the orbit with n = 2.

4. What are the allowed values for each of the four quantum numbers: n, l, ml, and ms?

5. State the Heisenberg uncertainty principle. Describe briefly what the principle implies.

6. Using complete subshell notation (1s22s22p6, and so forth), predict the electron configurations of the following ions.
(a) N3-
(b) Ca2+
(c) S-
(d) Cs2+
(e) Cr2+
(f) Gd3+

Verified Expert

The problems in this chapter are based in law of conservation of energy., which states that energy can neither be created nor be destroyed, but it can be changed from one form to another. There are different forms of energy and in this particular chapter, the emphasis is on heat energy. When an object which is hot is brought into contact with cold body and assuming there are no heat losses to the surroundings, heat energy is transferred from hot body to cold body until they reach an equilibrium temperature where there is no further heat transfer. Electromagnetic radiation deals with the frequencies and wave lengths of photons. If Frequency is known one can identify the type of radiation. Depending upon the wave length and frequency, the radiation is divided into X-ray, gamma ray etc. So one can identify the type of radiation depending upon the wave length.

Reference no: EM131462450

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