hmwk-6-solutions-Fall-2006

hmwk-6-solutions-Fall-2006 - Homework 6. Solutions Problem...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 6. Solutions Problem 1. d) If two objects are in thermal equilibrium they have the same temperature. Problem 2. e) In constructing a thermometer it is necessary to use a substance that undergoes some change when heated or cooled. Problem 3. b) For a constant volume gas thermometer p T = const. , therefore: p T = p T p = p- p = p ( T T- 1) = 6 . 3 10 4 Pa ( 274 . 16 K 273 . 16 K- 1) = 231 Pa Problem 4. b) The area of the face of a coin is A = r 2 = d 2 / 4. If a diameter increases by 0.17%, d 1 . 0017 d then A 1 . 0017 2 A , so the area will increase by 0.34%. Problem 5. b) Each side of the iron cube will expand due to heat: a = T a = 10- 5 / C 50 C 5 cm a = a + a = 5 . 0025 cm From this we find the new volume to be V = a 3 = 125 . 1875 cm 3 . At 10 C volume of the cube is V = a 3 = 125 cm 3 , so we see that the volume will increase due to heating by 0 . 1875 cm 3 Problem 6. a) Heat is energy transfered due to a temperature difference. Problem 7. b) Heat has the same unit as work. Problem 8. a) The heat capacity of an object is equal to the amount of heat energy needed to raise its temperature by 1 C . 1 Problem 9. d) The specific heat of a substance is equal to the amount of heat energy per unit mass needed to raise the temperature of the substance by 1 C . Problem 10. d) The heat capacity at constant volume and the heat capacity at constant pressure have dif- ferent values because the system does work at constant pressure, but not at constant volume. Problem 11. b) From the change in internal energy we can find the change in temperature: T = E int mc where m is the mass of the cube and c is the specific heat. The mass of the cube is m = Al a 3 = 2 . 7 H 2 O a 3 . T = E int 2 . 7 H 2 O a 3 . 217 c H 2 O T = 1 . 71 E int H 2 O a 3 c H 2 O The density of water is H 2 O = 1 g/cm 3 and specific heat of water is 1 cal/g C . Using that we find T = 10 C Problem 12. b) If the objects are thermally isolated then the amount of heat that one object will lose is equal to the amount of heat the other object will gain. C A ( T final- T A ) = C B ( T B- T final ) Solving for T final we find: C A T final + C B T final = C B T B + C A T A and finally: T final = C B T B + C A T A C A + C B Problem 13. c) Using the formula from previous problem and C B = 2 C A we find T final = C B T B + C A T A C A + C B = C A (2 T B + T A ) 3 C A = 400 K 2 Problem 14. a) Energy needed to melt down A is ML where M is the mass of A and L is latent heat of fusion. The total heat energy that A receives is C A ( T final- T A ) + ML and this is equal to the heat energy lost by B....
View Full Document

This note was uploaded on 03/22/2008 for the course PHY 317k taught by Professor Kopp during the Fall '07 term at University of Texas at Austin.

Page1 / 10

hmwk-6-solutions-Fall-2006 - Homework 6. Solutions Problem...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online