Physics II Prelims Reviewer

PHYSICS 2 PRELIMS REVIEWER (2nd Semester 2011-2012) TEMPERATURE FORMULAS: °F = 9 5 °C + 32 °R = 8 25 °C + 460 °F = 45 8 °R – 5111 2 °C = 5 9 (°F – 32) °C = 25 8 (°R – 460) °R = 8 45 [°F + 5111 2 ] K = 273 + C C° = 9 5 F° = K THERMOMETER SCALES REFERENC E POINTS CELSIUS FAHRENHE IT KELVIN RANKINE STEAM PT. 100 o C 212 o F 373 K 492 o R FREEZING PT. 0 o C 32 o F 273K 460 o R ABSOLUTE ZERO -273 o C 0 K SAMPLE PROBLEMS: 1. Convert: a.) -15 o F to KELVIN answer: 246.89 K b.) 303 K to CELSIUS and to FAHRENHEIT answer: 30 o C and 86 o F c.) -459 o F to CELSIUS answer -272.78 o C 2. A thermometer has 0 o A freezing point and 80 o A, the boiling point of water at normal pressure. Give the equation to change A to o C, A to o F, and A to K. CELSIUS FAHRENHEIT KELVIN 100 O C 212 O F 373 K C F K 0 O C 32 O F 273 K TEMPERATURE SCALE OF “A” 80 O A A 0 O A A to C* A to K* 80− 0 A− 0 = 100− 0 C− 0 80− 0 A− 0 = 373− 273 K− 273 A to F* 80− 0 A− 0 = 212− 32 F− 32 *Gamitin nyo lang yung law of proportionality between SCALE A and STANDARD SCALE para makuha nyo ang equivalent formula ng A to the other standard temperature scale. 3. The change in temperature of a body is 45 C o . How much is it in F o ? In K? 45C° x 9 5 F° 1C° = 81F° 45C° x 1K 1C° = 45K * F o and C o is different from o F and o C. F o and C o is a notation for the change in temperature. ADDITIONAL PROBLEMS 1. While vacationing in Italy, you see on local TV one summer morning that the temperature will rise from the current 18 o C to a high of 39 o C. What is the corresponding increase in the Fahrenheit temperature? °F = 9 5 °C + 32 °F = 9 5 (18) + 32 °F = 9 5 (39) + 32 °F = 64.4°F °F = 102.2°F 102.2 – 64.4 = 37.8 F ° 2. Two beakers of water, A and B, initially are at the same temperature. The temperature of the water in beaker A is increased 10 o F, and the temperature of the water in beaker B is increased 10 K. After those temperature changes, which beaker of water has the higher temperature? Explain. °C A = 5 9 (°F – 32) °C B = K – 273 °C A = 5 9 (10 – 32) °C B = 10 – 273 °C A = -12.22°C °C B = -263°  Beaker A increased -12.22 o C while beaker B increased -263 o C which means beaker A has a higher temperature than beaker B. 3. You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10 K. What is its temperature change in F o and C o ? -10K = -10°C -10°C x 9 5 F° 1C° = -18F° THERMAL EXPANSION Linear: L f = L o + ΔL L f = L o (1 + α ΔT) ΔL = α L o ΔT where α = coefficient of thermal expansion Area: x Δx ΔA = A f – A o ΔA = 2xΔx ΔA = 2x (x α ΔT) x ΔA = 2α A o ΔT Δx Volume: ΔV = 3α V o ΔT ΔV = β V o ΔT x x where β solid < β liquid Over Flow = (β liquid – β solid ) V o ΔT x THERMAL STRESS (Tensile or Compressive) S = F A = Y ∆L L O where: Y = Young’s Modulus of Elasticity ∆L L o = fractional change due to stress ∆L L o = α∆T = fractional change due to change in length COMPRESSION: T f > T o TENSION: T f < T o UNITS: F N dynes lb A m 2 cm 2 in 2 S N/m 2 dynes/cm 2 lb/in 2 or lb/ft 2 Y N/m 2 dynes/cm 2 lb/in 2 or lb/ft 2 EXAMPLE PROBLEMS: 1. The ends of a steel rod, exactly 0.2 in 2 in cross-sectional area are rigidly held between two fixed points at a temperature of 30 o C. When the temperature drops to 20 o C, find: a.) the stress developed at the ends of the rod b.) the pull in the rod Y steel = 33 x 10 6 lb/in 2 α steel = 11 x 10 -6 /C o a.) S = F A = Yα∆T S = (33 x 10 6 )(11 x 10 -6 )(-10) = -3630 lb in 2 b.) F = SA F = (-3630)(0.2) = -726lb 2. An aluminum baseball bat has a length of 86 cm at a temperature of 17 o C. When the temperature of the bat is raised, the bat lengthens by 1.6 x 10 -4 m. Determine the final temperature of the bat. ( α Al = 23 x 10 -6 /C o ) L o = 86 cm or 0.86 m T o = 17 o C ΔL = 1.6 x 10 -4 m ∆T = ∆L L o α T f – 17 = 1.6 x 10 -4 (0.86)(23 x 10 -6 ) T f = 25.089°C 3. The brass bar and the aluminum bar as shown are each attached to an immovable wall. At 28 °C, the air gap between the rod is 1.3 x 10 -3 m. At what temperature will the gap be closed? BRASS ALUMINUM 2m 1m α Br = 19 X 10 -6 /C° α Al = 23 X 10 -6 /C° ∆L Br + ∆L Al = 1.3 x 10 -3 m (α L o ΔT) Br + (α L o ΔT) Al = 1.3 x 10 -3 m [(19 X 10 -6 )(2)∆T + (23 X 10 -6 )(1)∆T = 1.3 x 10 -3 m ∆T = 21.31C° T f = 28 + 21.31 = 49.31C° 4. A copper cylinder is initially at 20.0 o C. At what temperature will its volume be 0.150% larger than it is at 20.0 o C? (β Cu = 5.1 x 10 -5 ) Let X o = the initial volume Let X o + 0.0015 X o = the final volume Let ∆V = 0.150%V o ΔV = βV o ΔT 1.5 x 10 -3 V o = (5.1 x 10 -5 )(V o )(T f – 20) T f = 49.41°C 5. A steel tank is completely filled with 2.8 m 3 of ethanol when both the tank and the ethanol are at temperature of 32.0°C. When the tank and its contents have cooled to 18°C, what additional volume of ethanol can be put into the tank? (β e = 75 x 10 -5 /C o β s = 3.6 x 10 -5 /C o ) ΔV s = V o β s ΔT = (2.8) (3.6 x 10 -5 ) (-14) ΔV s = - 0.00141 ΔV s = - 1.41 x 10 -3 m 3 ~ - 1.41 L ΔV e = V o β e ΔT = (2.8) (75 x 10 -5 ) (-14) ΔV e = - 0.0294 ΔV e = -2.94 x 10 -2 m 3 ~ -29.4 L 6. A machinist bores a hole of diameter 1.35 cm in a steel plate at a temperature of 25°C. What is the cross sectional area of the hole a) at 25°C b) when the temperature of the plate is increased to 175°C? Assume that the coefficient of linear expansion remains constant over this temperature range. a. A 0 = r 0 2 = (0.675) 2 =1.431 cm 2 b. ∆A = (2∝)A o ∆T = 2(1.2 x 10 -6 )(1.431)(175-25) = 5.15 x 10 -5 cm 2 A = A o + ∆A = 1.431 + (5.15 x 10 -6 ) = 1.436 cm 2 7. The outer diameter of a glass jar and the inner diameter of an iron lid are both 725 mm at room temperature (20°C). What will be the size of the mismatch between the lid and the jar if the lid is briefly held under hot water until its temperature rises to 50°C, without changing the temperature of the glass? *Find the change ΔL in the diameter of the lid. ΔL = αLΔT = (1.2.10− (C°)−)(725 mm)(30.0 C°) = 0.26 mm 8. An iron rod and a zinc rod have lengths of 25.55 cm and 25.50 cm respectively, at 32°F. At what temperature will the rods have the same lengths? The coefficients of linear expansion of iron and zinc are 0.000010 per C° and 0.000030 per C°, respectively. L 0 iron = 25.55 cm L 0 zinc = 25.50 cm T 0 = 32°F = 0°C T f = ? ∆T = T f - T o L iron = L zinc L o iron (1 + ∝ ∆T) = L o zinc (1 + ∝ ∆T) 25.55 [1 + (0.000010)(∆T)] = 25.50 [1 + (0.000030)(∆T) 25.55 + 0.0002555 ∆T = 25.5 + 0.000765 ∆T 0.0002555∆ – 0.000765 ∆ = 25.5 – 25.5 - 0.0005095 ∆ = 0.05 ∆ = 98.14 ° T f = 98.14 + 0°C T f = 98.14 °C 9. a) A wire that is 1.50 m long at 20°C is found to increase in length by 1.90 cm when warmed to 420°C. Compute its average coefficient of linear expansion for this temperature range. (b) The wire is stretched just tant (zero tension) at 420°C. Find the stress in the wire if it is cooled to 20°C without being allowed to contract. Young’s modulus for the wire is 2.0 x 10 11 Pa. (a) ΔL=αLΔT ∝ = ∆ L Lo∆T = 0.019 m 1.50m(420°C−20°C) =3.2 x 10 -5 (C°) -1 (b) Stress F/A= −YαΔT ΔT=20°C−420°C=−400 C° *ΔT always means final temperature minus initial temperature F/A= −(2.0.10 11 Pa)(3.2.10 -5 (C°) -1)(−400 C°) = +2.6.10 9 Pa 10. A glass flask whose volume is 1000 m 3 at 0°C is completely filled with mercury at this temperature. When flask and mercury are warmed to 55°C, 8.95 cm 3 of mercury overflow. If the coefficient of volume expansion of mercury is 18 x 10 -5 /K, compute the coefficient of volume expansion of the glass. ∆V Hg - ∆V glass = 8.95 cm 3 ∆V Hg = V o Hg ∆ = (1000) (18 x 10 -5 ) (55) = 9.9cm 3 ∆V glass = ∆V Hg - 8.95 = 9.9 – 8.95 = 0.95 glass = ∆Vglass ∆ = 0.95 1000 (55) = 1.7 x 10 -5 /K HEAT, Q Q = m c ΔT where: c = specific heat capacity UNITS: Q calorie Joule (kcal) Btu m gram Kg lb ΔT C o C o F o c cal/gC o J/kgC o or kcal/kgC Btu/lbF o CONVERSION FACTOR: 1 Btu = 252 cal 1 kcal = 1000 cal 1 cal = 4.186 J c water = 1 cal gC o = 1 kcal kgC o = 1 Btu lbF o CALORIMETRY (Principle of conservation energy) HEAT LOST by warmer bodies is equal to the HEAT GAINED by the cooler bodies. Q LOSS = Q GAINED SAMPLE PROBLEMS: 1. The Calorimeter cup, 0.15kg in mass, is made of aluminum and contains 0.2kg of water. Initially the water and the cup have a common temperature of 18 o C. A 40g mass of solid material is heated to a temperature of the water, the cup and the solid material was observed at 22 o C, after thermal equilibrium was established. Q LOSS = Q GAINED 40 c (97 – 22)C o = [C al m Al + c w m w ] (22 – 18) c = [ (200▪0.22) + (150▪1) ](4) (40)(75) c = 0.31 cal gC° 2. A substance boils at 120 o C and freezes at -20 o C. Its specific heat capacity as a solid, liquid, and gas are 1, 1.5, and 0.4 cal/gC o respectively. When 15g of this substance at 180 o C is brought to -40 o C, 62.74 Btu of heat is liberated. If the L f of the substance is 200 cal/g, determine its L v ? 62.74Btu x 252cal 1Btu = 15,810.48 cal Q = mc∆T + mL f + mc∆T + mL v + mc∆T 15810.48 = 15[-20-(-40)] + (15)(200) + (15)(1.5)[120-(- 20)] + 15L v + (15)(0.4)(180-120) 15810.48 = 6810 _ 15L v L v = 600 cal g 3. How much water could be cooled from 22°C to 5°C by a 10-g ice cube placed into a glass of water? The glass container is initially at 5° C and its water is equivalent is 10-g. *water equivalent means yung mass ng water is equal to mass of container. Equal din ang value of C Q lost = Q gained m ice C water (T f -T i ) + m ice L f = m water C water (T f— T i ) + m glass C water (T f -T i ) (10g)(1)(22°C - 0°C) + (10g)(80 cal/g) = m w (1)(22°C - 5°C) + (10g)(1)(22°C - 5°C) 220 + 800 = 17 m w + 170 17 m w = 850 m w = 50 g 4. When 200g of copper at 100°C is dropped into an aluminum container of mass 120g and containing 200g of an unknown liquid at 15°C, the mixture reaches a temperature of 45°C. What is the specific heat of the unknown liquid? C of copper is 0.093 and that of aluminum is 0.21. Q lost = Q gained m copper C copper (T f -T i ) = m al C al (T f -T i) + m iquid C liquid (T f -T i ) 200g(0.093)(100°C-45°C) = 120g(0.21)(45°C-15°C) + 200g(C liquid )(45°-15°C) 1023 = 756 + 6000 C liquid C liquid = 0.0445 5. A worker needs to know the temperature inside an oven. He removes a 2-lb iron bar from the oven and places it in a 1-lb aluminum container partially filled inside an oven with 2-lb of water. The system is immediately insulated and the temperature of the water and container rises from 70°F to the equilibrium temperature of 120°F. What was the oven temperature? C of iron = 0.113 and C of Al = 0.22 Q lost = Q gained m iron C iron ∆T = m wate C wate ∆T + m al C al ∆T (2)(0.113)(T - 120) = [(2)(1)(120 – 70) + (1)(0.22)(120 – 70)] -27.12 + 0.226 T = 100 + 11 0.226 T = 138.12 T = 611.15 °F 6. A substance boils at at 120°C and freezes at -20°C. Its specific heat as a solid, a liquid, and a gas are 1, 1.5 and 0.40 cal/g-C°, respectively. When 15g of this substance at 180°C is brought to -40°C, 62.738 of heat is liberated. If the heat of fusion of the substance is 200 cal/g, what is the heat of vaporization? *may tatlong∆T kasi 3 beses siya maguundergo ng temperature change. First, from -40° to -20°C.Second, from - 20°C to 120°C.Third, from 120°C to 180°C.maymLfkasimaguundergosiyang phase change from solid to liquid Lv from liquid to steam. Q = mc∆T + mL f + mc∆T + mL v +mc∆T = 15(1)(20) + 15(200) + 15(1.5)(140) + 15L v + 15(0.4)(60) = 5010 + L v L v = 600 cal/g HEAT TRANSFER I. CONDUCTION H = KA∆T L where: H = rate of heat flow t = time of heat flow K = thermal conductivity A = cross-sectional area L = thickness of material ∆T L = temperature gradient II. CONVECTION H = Q t = K L A∆T = h A∆T = A▪T 1 h f + L K + 1 h f where: h = film coefficient III. RADIATION P = eσAT 4 where: P = radiant power (Heat Rate) P = Q t e = emissivity of material A = surface area T = temperature of the material net radiant power: P = eσA(T 4 – T s 4 ) where: T s = surrounding temperature σ = Stefan – Boltzmann Constant σ =5.7 x 10 -8 W m 2 K 4 or 5.7 x 10 -12 W cm 2 K 4 *Note: for BLACKBODY, e = 1 SAMPLE PROBLEMS: 1. The temperature at the ends of a bar are 85 o C and 27 o C. The bar has a length of 68 cm. What is the temperature at a point 22 com from the cooler end of the bar? H 2T = H T1 KA∆T 2T L 2T = KA∆T T1 L T1 T 2 - T 68 - 22 = T - T 1 22 T = 45.76°c 2. Compute the heat conducted per hour through a plate- glass window of area 15 ft 2 and thickness ¼ in., when the inside temperature is 70 o F and the outside temperature is 20 o F. The thermal conductivity of the glass is 5.8 Btu/[h ft 2 (F o /in)] and the film coefficient for the fluid is about 2 Btu/(h ft 2 F o ). Q t = A∆T 1 h + L k + 1 h Q t = 15(70 - 20) 1 2 + 1 4 5.8 + 1 2 Q t = 719.01 Btu h in one day, find the amount of heat flow. H = Q t Q = Ht Q = 719.01 Btu h x 24h Q = 17,256.2 Btu 3. A spherical body emits 1 x 10 8 watts at 1000K surface temperature. Assuming the body is a perfect emitter, find its radius. P = eσAT 4 A = P eσT 4 A = 4πr 2 4πr 2 = P eσT 4 r = P 4πeσT 4 r = 1 x 10 8 4π(1)(5.6 x 10 -8 )(1000) 4 r = 11.85m 4. (17.66/page 604) One end of an insulated metal rod is maintained at 100.0˚C, and the other end is maintained at 0.00˚C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm 2 . The heat conducted by the rod melts 8.50 g of ice in 10.0min. Find the thermal conductivity k of the metal. Given: T H = 100.0˚C T C = 0.00˚C L = 60.0 cm = 0.600 m A = 1.25 cm 2 = 1.25 x 10 -4 m 2 m = 8.5 g = 0.0085 kg L f = 80 kcal/kg × 4186 1 = 3.34 × 10 5 Solution: Q = mL f = (0.0085 kg)( 3.34 × 10 5 ) Q= 2.84 × 10 3 = 2.84× 10 3 10 × 1 60 = 4.73 = = ( − ) = ( − ) = 4.73 ( 0.600 ) 1.25 ×10 −4 2 (100°) = 227 ∙ 5. (17.67/page 604) A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and layer of Styrofoam insulation 2.2 cm thick on the inside wall surface. The wood has k = 0.080 W/m·K , and the Styrofoam has k= 0.010 W/m·K. the interior surface temperature is 19.0℃. the exterior temperature is - 10.0℃. (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through the wall? Given: T 1 = -10˚C = 263 K T 2 =19˚C=292 K = 3.0 cm=0.03 m = 2.2cm = 0.022 m = 0.080 W/m·K = 0.010 W/m·K Solutions: (a) = ( − ) = ( − 1 ) = ( 2 − ) = − 1 = 2 − − 1 = 2 − (0.080 W/m·K) (0.022m) (T – 263)K = (0.010 W/m·K) (0.03m)( 292 – T)K T ·0.00176 W/K – 0.46288 W = 0.0876 - T·0.0003 W/K T·0.00206 W/K = 0.55048 W T = 267.22 K T = - 5.8 ˚C (b) = ( − ) = ( − ) Wood: = ( − 1 ) = (0.080 W/m·K) 267.22 – 263K 0.03m = 11.3 2 Styrofoam: = 2 − = (0.010 W/m·K) 292−267.22K 0.022m = 11.3 2 6. Heat Flow and Cooling your Coffee The author likes to bring a thermos bottle of coffee in his office each morning. The walls and the lid of a thermos bottle have low thermal conductivities so the rate of heat flow of the bottle is small. Suppose the coffee inside has a mass m= 0.30 kg and an initial temperature of 65˚C. Coffee is mostly water, so its specific heat is about the same as that of water. If the average rate of heat flow through the walls of the thermos bottle is Q/t = 3.0 W, approximately how long will it take the coffee to cool to a final temperature of 55˚C? Take a room temperature to be 25˚C. Given: = 0.30 = 65℃ = 3.0 / = 55℃ = 1 ∙ ° = 4186 ∙ Solution: = = ∆ = 0.30 (4186 ∙ )55 −65 = −1.3 × 10 4 *The negative sign indicates that the energy flows out of the coffee. = = 1.3 × 10 4 / = 4300 7. A small dog has thick fur, with a thermal conductivity of 0.040 W/(m _ K). The dog’s metabolism produces heat at a rate of 40 W and its internal (body) temperature is 38°C. If all of this heat flows out through the dog’s fur, what is the outside temperature? Assume the dog has a surface area of 0.50 m2, and that the length of the dog’s hair is 1.0 cm. Given: = 40 = 0.04 ∙ = 38℃ = 311 = 1.0 = 0,1 = 0.50 2 Solution: = ∆ 40 = 0.040 ∙ 0.50 2 311 − 0.01 = 291 = 291 −273 = 18℃ 8. A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice- water mixture at the other. The rod consists of a 1.00 m section of copper (one end in boiling water) joined end to end to a length L 2 of steel (one end in the ice-water mixture). Both reactions of the rod have cross-sectional areas of 4 cm 2 . The temperature of the copper-steel junction is 65°C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length L 2 of the steel section? For copper, c k=385 W/m⋅K. For steel, s k = 50.2 W/m⋅K. Q/t is the same for both sections of the rod. *For the copper = 385 0.0004(100−65) 1 = 5.39 J/s *For steel L = ∆ / = 50.2 0.0004(65−0) 5.39 / = . 9. Rods of copper, brass, aluminum and steel are welded together to form an X-shaped figure. Assume that the cross-sectional area of each rod is the same. The end of the copper rod is maintained at 120°C. The end of the aluminum rod is at 100°C and the ends of the brass and steel rods are at 0°C. The lengths and thermal conductivities of the rods are: copper, 50cm and 0.92; aluminum, 60cm and 0.49; brass, 25cm and 0.26; steel, 15cm and 0.12. Thermal conductivities are expressed in the CGS units. Calculate the temperature of the junction. ( ) Cu + ( ) Al = ( ) Br + ( ) St (0.92 (120 −))/50 + (0.49 (100 − ))/60 = (0.26 ( − 0))/25 + (0.12 ( − 0))/15 T = 67.26 °C 10. The emissivity of tungsten is 0.350. A tungsten sphere with radius 1.50 cm suspended within a large evacuated enclosure whose walls are at 290K. What power input is required to maintain the sphere at a temperature of 3000 K if heat conduction along the supports is neglected? H=Aeσ(T f 4 −T i 4 ). H=4π (0.0150 m) 2 (0.35)(5.67.10−8 W/m 2 ⋅K 4 )([3000 K] 4 −*290 K+ 4 ) = 4.54 x 10 4 W 11. A square chip of width L = 15 mm on a side are mounted to a substrate that is installed in an enclosure whose walls and air are maintained at a temperature of T surr = 25°C. The chips have an emissivity of 0.60 and a maximum allowable temperature of T s = 85°C. (a) If heat is neglected from the chips by radiation and natural convection, what is the maximum operating power of each chip? The convection coefficient depends on the chip-to-air temperature difference and may be approximated as h = C (T s – T surr ) 0.25 where C = 4.2 W/m 2 -K 5/4 (b)If a fan is used to maintain air flow through the enclosure and heat transfer is by forced convection, with h = 250 W/mK, what is the maximum operating power? SOLUTION: (a) The maximum operating chip power is the summation of heat transfer due to convection and radiations P max = Q tot = Q conv + Q rad = hA (T s – T surr ) + efA (T s 4 – T surr 4 ) = (4.2)(0.015) 2 (85 – 25) 1.25 + 5.67 x 10 -8 (0.6)(1)(0.015) 2 (358 4 -298 4 ) = 0.2232 W (b) Maximum operating power if a fan used is P max = Q tot = Q conv + Q rad = hA (T s – T surr ) + efA (T s 4 – T surr 4 ) = 250 (0.015) 2 (85 – 25) + 5.67x10 -8 (0.6)(1)(0.015 2 ) (358 4 – 298 4 ) = 3.44 W 12. What is rate of energy radiation per unit area of a blackbody at a temperature of (a) 273 K (b) 2730 K? *Remember that the e of blackbody is 1. = eT 4 (a) = (1)(5.67 x 10 -8 W/m 2 K 4 )(273 K) 4 = 315 W/m 2 (b) = (1)(5.67 x 10 -8 W/m 2 K 4 )(2730 K) 4 = 3.15 x 10 6 W/m 2 13. One end of a solid cylindrical copper rod 0.200 m long is maintained at a temperature of 20 K. The other end is blackbody and exposed to thermal radiation from surrounding walls at 500 K. The sides of the rod are insulated, so no energy is lost or gained except at the ends of the rod. When equilibrium is reached, what is the temperature of the blackened end? *Dahil blackened yung end, e = 1 so T 1 = 20 K and T 2 yunghahanapin. T s = 500 K T 2 = T 1 + (T s 2 – T 1 4 ) = T 1 + (6.79 x 10 -12 K -3 ) (T s 4 – T 1 4 ) = T 1 + 0.424 K = 20.42 K EXERCISES: 1. The melting point of tin is 232°C. What is this temperature on the Fahrenheit and Kelvin scales? 2. What is the value of the “absolute zero” of temperature on the Fahrenheit scale? 3. How much heat is required to raise the temperature of a 3.0-kg block of steel by 50 K? 4. While working in a research lab, you find that it takes 400 J of heat to increase the temperature of 0.12 kg of a material by 30 K. What is the specific heat of the material? 5. You are a blacksmith and have been working with 12 kg of steel. When you are finished shaping it, the steel is at a temperature of 400°C. To cool it off, you drop it into a bucket containing 5.0 kg of water at 0°C. How much of this water is converted to steam? Assume the steel, the water, and the steam all have the same final temperature. 6. One section of a steel railroad track is 25 m long. If its temperature increases by 25 K during the day, how much does the track expand? Cu Brass Al Steel 7. A steel building is 120 m tall when the outside temperature is 0°C. How tall is the building on a hot summer day (30°C)? The change in height is small, so express your answer with five significant figures. 8. A rectangular aluminum plate has an area of 0.40 m 2 at 15°C. If it is heated until its area has increased by 4.0 x 10 6 m 2 , what is the final temperature of the plate? 9. An aluminum beam 80 m long and with a cross-sectional area of 0.10 m 2 is used as part of a bridge. The beam is clamped rigidly at both ends. (a) If the temperature of the beam is increased by 30°C, what is the extra force exerted by the beam on one of its supports? (b) What is the force if the temperature of the beam is decreased by 40°C? 10. A small dog has thick fur, with a thermal conductivity of 0.040 W/(m K). The dog’s metabolism produces heat at a rate of 40 W and its internal (body) temperature is 38°C. If all of this heat flows out through the dog’s fur, what is the outside temperature? Assume the dog has a surface area of 0.50 m 2 , and that the length of the dog’s hair is 1.0 cm. 11. A jogger generates heat energy at a rate of 800 W. If all this energy is removed by sweating, how much water must evaporate from the jogger’s skin each hour? 12. A typical incandescent lightbulb has a filament temperature of approximately 3000 K. At what wavelength is the intensity of the emitted light highest? 13. The intensity of sunlight at the Earth’s surface is about 1000 W/m 2 . What is the total power emitted by the Sun? 14. An aluminum block slides along a horizontal surface. The block has an initial speed of 10 m/s and an initial temperature of 10°C. The block eventually slides to rest due to friction. Assuming all the initial kinetic energy is converted to heat energy and all this energy stays in the block, what is the final temperature of the block? 15. The Rankine scale. Named after William John Macquorn Rankine (a Scottish engineer and physicist who proposed it in 1859), the Rankine scale is similar to the Kelvin scale in that the zero point is placed at absolute zero, but the size of temperature differences are the same as that of the Fahrenheit scale (e.g., 1°F = 1°R). (a) Determine the conversion formula to go from the Fahrenheit to the Rankine scale. (b) Find the formula to convert from Kelvin to Rankine. (c) What is the temperature of the freezing point of water on the Rankine scale? What is room temperature on this scale? 16. A machinist bores a hole of diameter 1.35 cm in a steel plate at a temperature of 25°C. What is the cross sectional area of the hole a) at 25°C b) when the temperature of the plate is increased to 175°C? Assume that the coefficient of linear expansion remains constant over this temperature range. 17. The outer diameter of a glass jar and the inner diameter of an iron lid are both 725 mm at room temperature (20°C). What will be the size of the mismatch between the lid and the jar if the lid is briefly held under hot water until its temperature rises to 50°C, without changing the temperature of the glass? 18. An iron rod and a zinc rod have lengths of 25.55 cm and 25.50 cm respectively, at 32°F. At what temperature will the rods have the same lengths? The coefficients of linear expansion of iron and zinc are 0.000010 per C° and 0.000030 per C°, respectively. 19. A long rod, insulated to prevent heat losses, has one end immersed in boiling water and the other end in water-ice mixture all at atmospheric pressure. The rod consists of 100 cm of copper (one end in steam) and length, L 2 , of steel (one end in ice). Both rods have cross-sectional area of 5 cm 2 . The temperature of the Cu-steel junction is 60°C, K Cu = 0.92 and K steel = 0.12 cal/s-cm-C°. (a) How many calories per second flow from the steam bath to the ice water mixture? (b) How long is L 2 ? 20. One end of an insulated metal rod is maintained at 100°C and the other end at 0°C by an ice-water mixture. The rod has a length of 40cm and a cross-section of 0.750 cm 2 . The heat conducted by the rod melts 3.00 g of ice in 5.00 min. Calculate the thermal conductivity of the metal. 21. What is the final result when 400g of water and 100g of ice at 0°C are in a calorimeter whose water equivalent is 50g into which is passed 10 g of steam at 100°C? 22. A cylindrical hot plate is maintained at a temperature of 120°C by an electrical heating coil placed beneath it. Heat is being transferred the surrounding air through natural convection. The room temperatue is 30°C. Determine the diameter of the hot plate if the heating coil is penetrating heat at a rate of 800 W. Assume that the heat generated by the coil is totally effective in heating the plate. h for a horizontal place facing upward is 0.595 x 10 -4 () 1/4 23. A long rod, insulated to prevent heat losses, has one end immersed in boiling water and the other end in water-ice mixture all at atmospheric pressure. The rod consists of 100 cm of copper (one end in steam) and length, L 2 , of steel (one end in ice). Both rods have cross-sectional area of 5 cm 2 . The temperature of the Cu-steel junction is 60°C, K Cu = 0.92 and K steel = 0.12 cal/s-cm-C°. (a) How many calories per second flow from the steam bath to the ice water mixture? (b) How long is L 2 ? 24. One end of an insulated metal rod is maintained at 100°C and the other end at 0°C by an ice-water mixture. The rod has a length of 40cm and a cross-section of 0.750 cm 2 . The heat conducted by the rod melts 3.00 g of ice in 5.00 min. Calculate the thermal conductivity of the metal. 25. What is the final result when 400g of water and 100g of ice at 0°C are in a calorimeter whose water equivalent is 50g into which is passed 10 g of steam at 100°C? 26. A cylindrical hot plate is maintained at a temperature of 120°C by an electrical heating coil placed beneath it. Heat is being transferred the surrounding air through natural convection. The room temperatue is 30°C. Determine the diameter of the hot plate if the heating coil is penetrating heat at a rate of 800 W. Assume that the heat generated by the coil is totally effective in heating the plate. h for a horizontal place facing upward is 0.595 x 10 -4 (∆) 1/4 27. A tungsten filament reaches a temperature of 2000 K when its power consumption is 16 watts. On the assumption that the filament radiates heat as a blackbody, what power must be supplied to maintain a filament temperature of 3000 K? ANSWERS: 1. T F = 450°F and T K = 505 K 2. 460°F 3. 6.8 X 10 4 J 4. 110 J/(kg K) 5. 0.26 kg 6. 7.5 X 10 3 m 7. 120.04 m 8. 155°C 9. (a) 4.7 X 10 6 N (compression) (b) 6.3 X 10 6 N (tension) 10. 18°C 11. 1.3 kg/h 12. 9.6 X 10 7 m 13. 2.8 X 10 26 W 14. 10.06°C 15. (a) T R = T F + 459 (b) T R = (9/5)T K (c) Freezing point of water = 491°R; room temperature = 530°R 16. A 0 = 1.431 cm 2 A = 1.436 cm 2 17. ΔL = 0.26 mm 18. T f = 98.14 °C 19. 1.84 cal/s, 19.57 cm 20. 0.427 21. 0°C, 490 g H 2 O, 20g ice 22. 121.46 cm 23. 1.84 cal/s, 19.57 cm 24. 0.427 25. 0°C, 490 g H 2 O, 20g ice 26. 121.46 cm 27. 81 J/s