Thus, -195.75 °C, 77.4 K, and -320.35 °F are equivalent temperatures on dif-ferent scales,(b) For a change ATc=[-100°C-(-195.75 °C)]=95.75 c°, we use Eq. 11.3to find the change in temperature on the Fahrenheit scale as:ATF = ATC = ?I-100 - (-195.75)1 = 172.35 FThus, a change 95.75 C° = 172.35 F, where the notations C° and F refer totemperature difference, not to be confused with actual temperatures, which arewritten in terms of symbols °C and °F.(1) The melting point of gold is 10640C and the boiling point is 26600C. (a) Express thesetemperatures in Kelvins. (b) Compute the difference of these temperatures in Celsius degrees andKelvin degrees and compare the two numbers.(2) Liquid nitrogen has a boiling point of -195.810C at atmospheric pressure. Express this temperaturein (a) degrees Fahrenheit, and (b) kelvins.ANS. (a) -3210F (b) 77.30C(3) A substance is heated from -120F to 1500F. What is its change in temperature on (a) the Celsiusscale and (b) kelvin scale?ANS. (a) 900C (b) 90K(4) A certain process cools a body from 3500C to -800C. Express the change in temperature in (a)kelvins, and (b) Fahrenheit degrees.(5) At what temperature are the reading from a Fahrenheit thermometer and Celsius thermometer thesame?التمد د الحراري ل لمواد الصلبة والسائلThermal expansion of solids and Liquids. من المعلوم أن معظم الأجسام ت تمد د expandعندما تزداد درجة حرارتها. هذه الظاهرة تلعب دورأرئيسيا في العديد من التطبيقات الهندسية، فعلى سبيلالمثال يتم ترك مسافات بين الوصلات الحديدية فيالمباني والجسور والسك ك الحديدية والطرق السريعةالتعطى المجال ل لتمد د والإنكماش وإذا لم يتم فعل ذلكيمكن أن يتصدع المبني أو تنهار الجسور وتلتويالسك ك الحديدية بفعل التمد د الحراري ل لمواد المصنوعةsola sill ülthermal expansionل لاجسام هو نتيجة عن ل لتغير الذي يحدث ل لمسافاتالفاصلة بين جزيئات وذرات المادة. ولفهم أدق لماذكرناه لنظر إلى الشكل الموضح أدناه حيث يعبر عنالتركيب البلوري لمادة في الحالة الصلبة والتي تحتويعلى مصفوفة مرتبة من الذرات المترابطة مع بعضهاالبعض بفعل القوى الكهربية )الزنبرك في الشكل يمثلالقوى الكهربية(.eeeeeatيمثل الشكل بلورة لمادة صلبةوالزنبرك بين الذرات يمثلالقوى الكهربية التي تربطالذرات بعضها ب بعضعند درجات الحرارة العادية ت تذبذب الذرات حول موضع استقرارها في البلورة وتبلغ

Answered: Image /qna-images/answer/7d014c48-e68a-43fe-9531-fde96773ae37.jpg

(1) The melting point of gold is 10640C and the boiling point is 26600C. (a) Express these temperatures in Kelvins. (b) Compute the difference of these temperatures in Celsius degrees and Kelvin degrees and compare the two numbers.

(1) The melting point of gold is 10640C and the boiling point is 26600C. (a) Express these temperatures in Kelvins. (b) Compute the difference of these temperatures in Celsius degrees and Kelvin degrees and compare the two numbers.

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter1: Temperature And Heat

Section: Chapter Questions

Problem 119AP: For the human body, what is the rate of heat transfer by conduction through the body's tissue with...

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