521_PartUniversity Physics Solution

521_PartUniversity Physics Solution - Temperature and Heat...

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Temperature and Heat 17-7 17.43. IDENTIFY: . Qm cT The mass of n moles is . mn M = SET UP: For iron, 3 55.845 10 kg/mol M and 470 J/kg K. c = EXECUTE: (a) The mass of 3.00 mol is 3 (3.00 mol)(55.845 10 kg/mol) 0.1675 kg. M == × = [ ] / (8950 J)/ (0.1675 kg)(470 J/kg K) 114 K 114 C . TQ m c Δ= = ⋅ = = ° (b) For 3.00 kg, / 6.35 C . mT Q m c = = ° EVALUATE: (c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. 17.44. IDENTIFY: The latent heat of fusion f L is defined by f L = for the solid liquid phase transition. For a temperature change, . SET UP: At 1 min t = the sample is at its melting point and at 2.5 min t = all the sample has melted. EXECUTE: (a) It takes 1.5 min for all the sample to melt once its melting point is reached and the heat input during this time interval is 34 (1.5 min)(10.0 10 J/min) 1.50 10 J. ×= × f . L = 4 4 f 1.50 10 J 3.00 10 J/kg. 0.500 kg Q L m × = × (b) The liquid’s temperature rises 30 C ° in 1.5 min. . = Δ 4 3 liquid 1.00 10 J/kg K. (0.500 kg)(30 C ) Q c × = × Δ ° The solid’s temperature rises 15 C ° in 1.0 min. 4 3 solid 1.00 10 J 1.33 10 J/kg K. (0.500 kg)(15 C ) Q c × = × Δ ° EVALUATE: The specific heat capacities for the liquid and solid states are different. The values of c and f L that we calculated are within the range of values in Tables 17.3 and 17.4. 17.45. IDENTIFY and SET UP: Heat comes out of the metal and into the water. The final temperature is in the range 01 0 0 C , T << ° so there are no phase changes. system 0. Q = (a) EXECUTE: water metal 0 QQ += water water water metal metal metal 0 mc T Δ+ metal (1.00 kg)(4190 J/kg K)(2.0 C ) (0.500 kg)( )( 78.0 C ) 0 c ⋅° + ° = metal 215 J/kg K c =⋅ (b) EVALUATE: Water has a larger specific heat capacity so stores more heat per degree of temperature change. (c) If some heat went into the styrofoam then metal Q should actually be larger than in part (a), so the true metal c is larger than we calculated; the value we calculated would be smaller than the true value. 17.46. IDENTIFY: Apply to each object. The net heat flow system Q for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the &±water² by w. T is the final equilibrium temperature. w 4190 J/kg K. c = KC . TT Δ EXECUTE: (a) system 0 Q = gives mm m ww w 0. mC T Δ = mm m ww w () 0 . mCTT mCTT +− = w . mC T T −= Solving for T , mmm www . mCT T mC + = + (70.0 kg) (3480 J/kg K) (37.0 C) (0.355 kg) (4190 J/kg C ) (12.0 C) 36.85 C (70.0 kg)(3480 J/kg C ) (0.355 kg) (4190 J/kg C ) T + ° ° ⋅°+ ° ° (b) It is possible a sensitive digital thermometer could measure this change since they can read to 0.1 C. ° It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. EVALUATE: Heat comes out of the body and its temperature falls. Heat goes into the soft drink and its temperature rises. 17.47. IDENTIFY: For the man’s body, . SET UP: From Exercise 17.46, 0 15 C T Δ=. ° when the body returns to 37 0 C .°. EXECUTE: The rate of heat loss is Q/t .

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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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521_PartUniversity Physics Solution - Temperature and Heat...

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