Tutorial 10 Solutions

Tutorial 10 Solutions - TUTORIAL#10 November 16-20th 2009...

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CHEM 1A03 TUTORIAL #10 November 16-20 th , 2009 Solutions – Chapter 7 ______________________________________________________________________________ 1. A bird has a much lower heat capacity than a person, and a much lower fraction of their body mass is water. A bird of mass 75 g, is caught out in a November rain, and its body temperature drops by 3 degrees. The bird generates energy through cellular respiration while flying, at a rate of 250 J /kilometer of flight. How far will the bird have to travel to raise his body temperature by 3.0 degrees, if his body’s heat capacity is 3.0 J/g. o C. J C C g J g q T C m q o o 675 3 . 0 . 3 75 required energy = 675 J Flight requirement = km km J J 7 . 2 / 250 675 2. (Ch 7, Q. 6). A piece of stainless steel (specific heat = 0.50 J.g -1 . o C -1 ) is transferred from an oven at 183 o C to 125ml of water, at 23.2 o C. The water temperature rises to 51.5 o C. What is the mass of the steel? Heat lost by steel = heat gained by water q steel = -(m x C x T) = q water = (m x C x T)  g m J m C C g J ml g ml m C C g J m o o o o 2 4 4 10 2 . 2 8 . 65 10 48 . 1 10 48 . 1 8 . 65 2 . 23 5 . 51 . 18 . 4 1 00 . 1 125 8 . 65 183 5 . 51 . 50 . 0 3. (Ch 7. Q. 14) Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of 1.00 mol of each substance, including the enthalpy change, H, for the reaction.
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Tutorial 10 Solutions - TUTORIAL#10 November 16-20th 2009...

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