1.- What is the in internal energy when a system releases 75 Jof heat energy and

Question

1.- What is the in internal energy when a system releases 75 Jof heat energy and does 54 J of work on its surroundings? a)21 J, b) -21 J, c) 129J, d) -1.4 J, e) -129 J 2.- A 275-g sample of nickel at 100 o C is placedin 100 mL of water at 22o C. What is the finaltemperature of the water? Assume that no heat is lost to or gainedfrom the surroundings. Specific heat capacity of nickel 0.444J/(g.K). 3.- How many calories are in 756.5 kJ(1 cal = 4.184 J)? 1.- What is the in internal energy when a system releases 75 Jof heat energy and does 54 J of work on its surroundings? a)21 J, b) -21 J, c) 129J, d) -1.4 J, e) -129 J 2.- A 275-g sample of nickel at 100 o C is placedin 100 mL of water at 22o C. What is the finaltemperature of the water? Assume that no heat is lost to or gainedfrom the surroundings. Specific heat capacity of nickel 0.444J/(g.K). 3.- How many calories are in 756.5 kJ(1 cal = 4.184 J)?

Explanation / Answer

( 1 ) Heat Energy is releases means q is taken as -ve& q = - 75 J
       Work is done on itssurroundings, so it is also taken as -ve & W = -54 J According to First Law of Thermodynamics ,                                                                      E = q + W                                                                            = -75 + ( -54 )                                                                            = - 129 J Change in internal Energy E is - 129 J ( 2 )  A 275-g sample of nickel at 100 oC is placed in 100 mL of water at 22o C. What is thefinal temperature of the water? Assume that no heat is lost to orgained from the surroundings. Specific heat capacity of nickel0.444 J/(g.K). Given initial temp of nickel t = 100 oC Initial temp of the water t ' = 22 oC Mass of water M = density of water * volume of thewater                            = 1 g / cm ^ 3 * 100 mL                            = 1 g / cm ^ 3 * 100 cm ^ 3                            = 100 g Let the final temperature be T Heat lost by Nickel = Heat gained by water                     mcdt = MCdT m = mass of Nickel = 275 g c = Specific heat capacity of nickel = 0.444 J/(g.K) dt = t-T     = 100-T C = Specific heat of water = 4.186 J / g oC dT = T - t '      =T -22 substitue values we get 275 ( 0.444) ( 100-T ) = 100( 4.186 ) ( T -2 )         12210 -122.1 T =418.6T-837.2                    540.7 T = 13047.2                              T = 24.13 oC ( 3 ) 756.5 KJ = 756.5 * 1000 J     Since1 KJ = 1000 J                      = ( 756.5 * 1000 ) / 4.184 cal                       ( Since 1 cal = 4.184 J )                      = 180807.839 cal

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