#2, solar, 22, 27, 33, gasses, 51

Ch 20.1 #2 Consider Joule's apparatus. The mass of each of the two blocks is 1.50 kg, and the insulated tank is filled with 200 g of water. What is the increase in the temperature of the water after the blocks fall through a distance of 3.00 m? ans 0.105 °C
We know Q = 0 in Joules apparatus (insulated) ΔEint = Q + W; ΔEint = W Work = 2 blocks (mgh) Work = 2(1.5*9.8*3)	Work = 88.2 J ΔEint = 88.2 J = mcDT 88.2 J = 0.2*4186*DT DT = 0.105 C°

Ch 20.2 #solar

We are going to heat the water in a calorimeter cup using solar power. Our solar collector has an area of 10.00 m², and the intensity delivered by sunlight is 600 W/m², how long does it take to increase the temperature of 300 cc of water from 20.0°C to 100.0°C? ans: t = 16.8 sec

m = ρ V m = 1 g/cc (300cc) m = 300 grams

Q = m c DT Q = 300g (1) (100-20) Q = 24 kilocalories (4186J = 1 kCal) Q = 100,000 J

P = Q / t (600W/m^2)(10m^2) = 100000J / t t = 16.8 sec

Ch 20.3 #22

Two speeding lead bullets, each of mass 5.00 g, and at temperature 20.0°C, collide head-on at speeds of 500 m/s each. Assuming a perfectly inelastic collision and no loss of energy by heat to the atmosphere, describe the final state of the two-bullet system. Ans 805 °C

Must add the kinetic energy of the two bullets Kone = ½ m v^2 Kone = ½ 0.005 (500)^2 Kone = 625 Joules Ktwo = 1250 Joules cPb = 128 J/kg Lf-Pb = 2.45 x 10^4 J/kg

Lead melts at 327 °C Qto melt point = m c DT Q = (2*0.005) (128) (327-20) Qto melt point = 393 Joules Energy to melt lead Qfusion = m Lf Qfusion = 0.01 (2.45 x 10^4) Qfusion = 245 Joules

Qremaining = m c DT (1250-393-245) = 0.01(128) DT DT = 478 °C So final temp is 327 °C + 478 C° T = 805 °C

Ch 20.4 #27

One mole of an ideal gas is heated slowly so that it goes from the PV state (P₀, V₀), to (3P₀, 3V₀), in such a way that the pressure is directly proportional to the volume. (a) How much work is done on the gas in the process? (b) How is the temperature of the gas related to its volume during this process?

During the heating process; P is proportional to V (given in the problem), so P = (constant) V where the constant = Pi/Vi which yields P = (Pi/Vi) V. Using same logic…if P is inversely prop to V then P = (Vi/Pi) 1/V, which is the normal situation (PiVi = PfVf).
(a) W = -∫ P dV W = -∫(Pi/Vi) V dV from Vi to 3Vi W = -(Pi/Vi) ½ (Vf^2 – Vi^2) from Vi to 3Vi W = -½(Pi/Vi) (9Vi^2 – Vi^2) W = -4 Pi Vi	(b) P V = nRT (Pi/Vi)V V = nRT T = (Pi / VinRT) V^2 Temperature must be proportional to the square of volume, rising to nine times its original value.

Ch 20.5 #33 A sample of an ideal gas is in a vertical cylinder fitted with a piston. As 5.79 kJ of energy is transferred to the gas by heat to raise its temperature, the weight on the piston is adjusted so that the state of the gas changes from point A to point B along the semicircle shown in Figure P20.33. Find the change in internal energy of the gas.

The work on the gas, W = -∫P dV, which is the area under the half circle curve plus the area below the half circle. Work = -½ π(height/2)(width/2) + 300kPa (4.8 liters) Work = -½ π(200kPa)(4.8 liters/2) + 300kPa(4.8 liters) Work = -2194 kPa*liters Work = -2194 Joules	We know that the area of a circle is πr^2. Area of a square is length * height. If we circumscribe a square about a circle of radius, r, then r = height/2 or r = width/2So the area of a circle can also be written as π(height/2)(width/2)
	∆Eint = Q + Work ∆Eint = 5790 J + -2194 J ∆Eint = 3596 Joules

Alternate method

A circle is % of a circumscribed square

π(h/2)² / h x h (h = w in a square)

π/4 = 78.5 %

(78.5% * 4.8 l * 200 kPa) + (300*4.8) = 2194 J

Ch 20.6 gasses

(a) Calculate the work done by the gas during process 1 à 2 (b) Calculate the work done by the gas during process 3 à 4 (c) Calculate the work done by the gas during process 1 à 2 à 3 à 4

(d) Calculate the work done by the gas during process 1 à 4 (e) Calculate the work done by the gas during process 4 à 5 à 2 à 3 (f) Calculate the work done by the gas during process 1 à 2 à 5 à 4 à 1

Calculate the work done by the gas during process 1 à 2 5P0 * (5-2)V0 = 15P0 V0 Calculate the work done by the gas during process 3 à 4 3P0 * (2-5)V0 = -9P0 V0 Calculate the work done by the gas during process 1 à 2 à 3 à 4 15P0 V0 + -9P0 V0 = 6P0 V0

Calculate the work done by the gas during process 1 à 4 5P0 * (0)V0 = 0 Calculate the work done by the gas during process 4 à 5 à 2 à 3 ½ b h + width h ½(5-2)V0 (5-3)P0 + (5-2)V0 3P0 12P0 V0

Calculate the work done by the gas during process 1 à 2 à 5 à 4 à 1 15P0 V0 + 0 + -9P0 V0 + 0 = 6P0 V0

Ch 20.7 #51

The intensity of solar radiation reaching the top of the Earth’s atmosphere is 1340 W/m². The temperature of the Earth is affected by the so-called greenhouse effect of the atmosphere. The result is that our planet’s emissivity for visible light is higher than its emissivity for infrared light. For comparison, consider a spherical object with no atmosphere, at the same distance from the Sun as the Earth. Assume its emissivity is the same for all kinds of electromagnetic waves and that its temperature is uniform over its surface. Identify the projected area over which it absorbs sunlight and the surface area over which it radiates. Compute its equilibrium temperature. Chilly, isn’t it? Your calculation applies to:

the average temperature of the Moon

astronauts in mortal danger aboard the crippled Apollo 13 spacecraft

global catastrophe on the Earth if widespread fires caused a layer of soot to accumulate throughout the upper atmosphere, so that most of the radiation from the Sun were absorbed there rather than at the surface below the atmosphere.

Our face showing to the sun is a circle…πr^2. We’re radiating out throughout the entire surface area of the planet 4πr^2. Stefan’s Law: P = σAeT^4 ; σ = 5.669 x 10^-8 W/m^2K^4

Pin = Pout σAineT^4 = σ Aout eT^4 πr^2e(1340) = σ(4πr^2)eT^4 (1340) = σ(4)T^4 T = 277 K or 4 °C