HW 4 - Phy 202

HW 4: Phy 202: Temperature, thermal Expansion and Heat: online 9 th version of the text book. Chapter 12; Page 381 Problems: Page 412 Questions 1, 2, 3, 4, 5; 10, 11, 12, 15, 17; 43, 46. Picture 1 – Screen shot of the page from the textbook to see HW problems from 1-5 and 10-17.

Picture 2 – Screen shot of the page from textbook to see Q 43 and 46. © 2021 Dr Dipti Sharma, screen shot of the text book pages 1. Suppose you are hiking down the Grand Canyon. At the top, the temperature early in the morning is a cool 3°C. By late afternoon, the temperature at the bottom of the canyon has warmed to a sweltering 34°C. What is the difference between and lower temperatures in (a) Fahrenheit degrees and (b) Kelvins? a) 1 F° = °C × 9/5 + 32 F° = 34°C × 9/5 + 32 F° = 3°C × 9/5 + 32 F° = 93.2 F° F° = 37.4 F° 34°C = 93.2 F° 3°C = 37.4 F°

3 – On the moon the surface temperature ranges from 375 K during the day to 1 × 10 2 K at night. What are these temperatures on the (a) Celsius and (b) Fahrenheit scales? a) °C = K – 273.15 T day = 375K T Night = 100K T day = 375 – 273.15 T Night = 100 – 273.15 T day = 101.85°C T Night = - 173.15°C b) °F = (K – 273.15) × 9/5 + 32 T day = 375K T Night = 100K T day = (375 – 273.15) × 9/5 + 32 T Night = (100 – 273.15) × 9/5 + 32 T day = 215.33 F° T Night = - 279.67 F° 4 – What’s your normal body temperature? It may not be 98.6°F, the often-quoted average that was determined in the nineteenth century. A more recent study has reported an average temperature of 98.2°F. What is the difference between these averages, expressed in Celsius degrees?

°C = (F° - 32) ÷ 9/5 °C = (98.6 - 32) ÷ 9/5 °C = (98.2 - 32) ÷ 9/5 °C = 37 °C = 36.78 98.6 F° = 37 °C 98.2 F° = 36.78 °C 37°C – 36.78°C = 0.22 °C 5 – Dermatologists often remove small precancerous skin lesions by freezing them quickly with liquid nitrogen, which has a temperature of 77 K. What is this temperature on the (a) Celsius and (b) Fahrenheit scales? a) °C = K – 273.15 °C = 77 – 273.15 77 K = - 196.15 °C b) °F = (K – 273.15) × 9/5 + 32 °F = (77 – 273.15) × 9/5 + 32 77 K = - 321.07 F°

ΔL = (12 × 10 -6 ) × 370 × (21 - 2) ΔL = 0.08436 m or 8.436 × 10 -2 m 12 – The Eiffel Tower is a steel structure whose height increases by 19.4 cm when the temperature changes from -9 to + 41° C. What is the approximate height (in meters) at the lower temperature? ΔL = α L 0 ΔT or L 0 = ΔL ÷ (α ΔT) ΔL = 0.194 m T 1 = - 9°C T 2 = 41°C α = 12 × 10 -6 °C -1 L 0 = ? L 0 = 0.194 ÷ {(12 × 10 -6 ) × [41 – (- 9)]} L 0 = 323.3 m or 3.233 × 10 2 m 15 – When the temperature of a coin is raised by 75°C, the coin’s diameter increases by 2.3 × 10 -5 m. If the original diameter of the coin is 1.8 × 10 -2 m, find the coefficient of linear expansion. ΔL = α L 0 ΔT or α = Δd ÷ (d 0 ΔT) Δd = 2.3 × 10 -5 m

ΔT = 75°C d 0 = 1.8 × 10 -2 m α = ? α = (2.3 × 10 -5 ) ÷ [(1.8 × 10 -2 ) × 75] α = 1.7 × 10 -5 °C -1 17 – One rod is made from lead and another from quartz. The rods are heated and experience the same change in temperature. The change in length of each rod is the same. If the initial length of the lead rod is 0.10 m, what is the initial length of the quartz rod? If ΔL lead = ΔL quartz then α L 0 ΔT = α L 0 ΔT Lead: Quartz: ΔL lead = ΔL quartz ΔL quartz = ΔL lead L 0lead = 0.10 m L 0quartz = ? ΔT lead = ΔT quartz ΔT quartz = ΔT lead α lead = 29 × 10 -6 °C -1 α quartz = 0.5 × 10 -6 °C -1 (29 × 10 -6 ) × (0.10) ΔT = (0.50 × 10 -6 ) × (L 0quartz ) ΔT Divide both sides by ΔT (29 × 10 -6 ) × (0.10) = (0.5 × 10 -6 ) × (L 0quartz ) L 0quartz = [(29 × 10 -6 ) × (0.10)] ÷ (0.5 × 10 -6 ) L 0quartz = 5.8 m

Q g = Q l Q l = Q g m g = m l m l = m g c g = 840 J/(kg×C°) c l = ? ΔT g = 83 – 53 = 30°C ΔT l = 53 – 43 = 10°C m g c g ΔT g = m l c l ΔT l (840) × 30 = (c l ) × (10) c = 2520 J/(kg×C°)