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# s_chapter_18 - Chapter 18 Heat and the First Law of...

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1685 Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 1 Object A has a mass that is twice the mass of object B and object A has a specific heat that is twice the specific heat of object B. If equal amounts of heat are transferred to these objects, how do the subsequent changes in their temperatures compare? ( a ) AB 4 Δ= Δ TT , ( b ) 2 Δ =Δ , ( c ) Δ , ( d ) B 2 1 A T T Δ = Δ , ( e ) B 4 1 A T T Δ = Δ Picture the Problem We can use the relationship T mc Q Δ = to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each. Express the change in temperature of object A in terms of its mass, specific heat, and the amount of heat supplied to it: A A A c m Q T = Δ Express the change in temperature of object B in terms of its mass, specific heat, and the amount of heat supplied to it: B B B c m Q T = Δ Divide the second of these equations by the first to obtain: B B A A A B c m c m T T = Δ Δ Substitute and simplify to obtain: ( )( ) 4 2 2 B B B B A B = = Δ Δ c m c m T T B 4 1 A Δ Δ T T = and ( ) e is correct. 2 Object A has a mass that is twice the mass of object B. The temperature change of object A is equal to the temperature change of object B when the objects absorb equal amounts of heat. It follows that their specific heats are related by ( a ) 2 = cc , ( b ) 2 = , ( c ) = , ( d ) none of the above Picture the Problem We can use the relationship T mc Q Δ = to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each.

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Chapter 18 1686 Relate the temperature change of object A to its specific heat and mass: A A A Δ c m Q T = Relate the temperature change of object B to its specific heat and mass: B B B Δ c m Q T = Equate the temperature changes and simplify to obtain: A A B B 1 1 c m c m = Solve for c A : B 2 1 B B B B A B A 2 c c m m c m m c = = = and ) ( b is correct. 3 [SSM] The specific heat of aluminum is more than twice the specific heat of copper. A block of copper and a block of aluminum have the same mass and temperature (20ºC). The blocks are simultaneously dropped into a single calorimeter containing water at 40ºC. Which statement is true when thermal equilibrium is reached? ( a ) The aluminum block is at a higher temperature than the copper block. ( b ) The aluminum block has absorbed less energy than the copper block. ( c ) The aluminum block has absorbed more energy than the copper block. ( d ) Both ( a ) and ( c ) are correct statements. Picture the Problem We can use the relationship T mc Q Δ = to relate the amount of energy absorbed by the aluminum and copper blocks to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum block: T c m Q Δ = Al Al Al Express the energy absorbed by the copper block: T c m Q Δ = Cu Cu Cu Divide the second of these equations by the first to obtain: T c m T c m Q Q Δ Δ = Al Al Cu Cu Al Cu Because the block’s masses are the same and they experience the same change in temperature: 1 Al Cu Al Cu < = c c Q Q or Al Cu Q Q < and ) ( c is correct.
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## This note was uploaded on 01/29/2012 for the course PHYS 213 taught by Professor Oshea during the Fall '08 term at Kansas State University.

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s_chapter_18 - Chapter 18 Heat and the First Law of...

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