Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 17 | From Eq. a 9 5 - 32 . b 9 5 32 . c 9 5 32 . From Eq. a 5 - 32 . b 5 9 107 - 32 . c 5 9 -18 - 32 . 1Co fFo so Fo T2 T1 Fo a 5 9 - C. b 5 9 - 44 C. a FromEq. 9 5 32 which is cause for worry. b 9 5X12 32 or 54oF to two figures. 9 5 Fo 1K 1Co po so a temperature increase of 10 K corresponds to an increase of 18 Fo . Beaker B has the higher temperature. For b ATC ATK Co. Then for a ATF ATC Co . Combining Eq. and Eq. Tk 5 Tf -32o and substitution of the given Fahrenheit temperatures gives a K b K c K. In these calculations extra figures were kept in the intermediate calculations to arrive at the numerical results. a TC 400 - 127oC TF 9 5 32 260oF. b TC 95 - -178oC TF 9 5 32 -289oF. c TC x 107 - x 107oC TF 9 5 x 107 32 x 107oF. From Eq. TK . From Eq. 1769OC. From Eq. 373 15K 444 mm. 1 x x On the Kelvin scale the triple point is K so OR 9 5 K . One could also look at Figure and note that the Fahrenheit scale extends from - 460OF to 32OF and conclude that the triple point is about 492or. From the point-slope formula for a straight line or linear regression which while perhaps not appropriate may be convenient for some calculators - x 10 Pa-------- 104 Pa which is 282OC to three figures. b Equation was not obeyed precisely. If it were the pressure at the triple nnint u Aiilrl P-iTP 1 AM Pa _ A nt. v 1 A4 p point would be p 2 37375 j 10 Pa. AT AL aL0 25 x 10-2 m 10-5 co - 1 m 168O C so the temperature is 183 o C. aL0AT x
