Urea, (NH 2 ) 2 CO, is used in themanufacture of resins and glues. When 5.00g of

Question

Urea, (NH2)2CO, is used in themanufacture of resins and glues. When 5.00g of urea isdissolved in 250.0 mL of water (d=1.00 g/mL) at 30oC ina coffee-cup calorimeter, 27.6 kJ of heat is absorbed. a) Is the solution process exothermic? b) What is qH2O? c) What is the final temperature of the solution? (Specificheat of the water is 4.18 J/g* oC) d) What are the initial and final temperatures inoF? The answers are a) no, b) -27.6 kJ, c) 3.6 oC, andd) ti = 86.0 oF and tf = 38.5oF. The one I seem to be having the most problems with is part c Idont understand how that value is obtained and any help would begreatly appreciated. Urea, (NH2)2CO, is used in themanufacture of resins and glues. When 5.00g of urea isdissolved in 250.0 mL of water (d=1.00 g/mL) at 30oC ina coffee-cup calorimeter, 27.6 kJ of heat is absorbed. a) Is the solution process exothermic? b) What is qH2O? c) What is the final temperature of the solution? (Specificheat of the water is 4.18 J/g* oC) d) What are the initial and final temperatures inoF? The answers are a) no, b) -27.6 kJ, c) 3.6 oC, andd) ti = 86.0 oF and tf = 38.5oF. The one I seem to be having the most problems with is part c Idont understand how that value is obtained and any help would begreatly appreciated. c) What is the final temperature of the solution? (Specificheat of the water is 4.18 J/g* oC) d) What are the initial and final temperatures inoF? The answers are a) no, b) -27.6 kJ, c) 3.6 oC, andd) ti = 86.0 oF and tf = 38.5oF. The one I seem to be having the most problems with is part c Idont understand how that value is obtained and any help would begreatly appreciated.

Explanation / Answer

Given that
5.00g of urea is dissolved in 250.0 mL of water (d=1.00 g/mL)at 30oC in a coffee-cup calorimeter, 27.6 kJ of heat isabsorbed. a) Heat is absorbed, hence the reaction isendothermic. b) qsol = +27.6 kJ    qH2O = - qsol                 =-27.6 kJ c)       qH2O = m s T        -27600 J = 250 g (4.18 J/g 0C) ( Tf - 300C)                   Tf = 3.6 0C d) We know that relation between 0C and0F                            0F = 0C*1.8 + 32                              Ti=30*1.8 + 32                                 =86 0F                             Tf = 3.6 *1.8 + 32                                 = 38.48 0F                             Tf = 3.6 *1.8 + 32                                 = 38.48 0F

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