Phthalates used as plasticizers in rubber and plastic products are believed to a

Question

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of delta H comb for dimethylphthalate (C10H10O4) is 4685 kJ/mol. Assume 0.805 g of Dimethylphthalate is combusted in a calorimeter whose heat capacity (C calorimeter) is 7.824 kJ/degree C at 21.295 degree C. What is the final temperature of the calorimeter? Calculate delta E for a system that absorbs 665.0 kJ of heat from its surroundings and does 446.0 kJ of work on its surroundings. How much heat is needed to raise the temperature of 115.0 g of water from 39 degree C to 95 degree C?

Explanation / Answer

given

dHcomb = 4685 kJ/mol

given

0.805 g of dimethylpthalate

we know that

moles = mass / molar mass

so

moles of dimethylpthalate = 0.805 / 194 = 4.15 x 10-3

now

dH = moles x dHcomb per mole

so

dH = 4.15 x 10-3 x 4685

dH = 19.44 kJ

this heat is used to raise the temp of the calorimeter

we know that

Heat = heat capacity x temp change

given

Heat = 19.44 kJ

heat capacity = 7.824 kJ

initial temp = 21.295 C

so

19.44 = 7.824 (Tf - 21.295)

Tf = 23.78

so the final temperature of the calorimter is 23.78 C

2)

we know that

Q = dE + W

given

Q = + 665 kJ

W = 446 kJ

so

665 = dE + 446

dE = 219 kJ

so

the dE for the system is 219 kJ

3) we know that

Heat = mass x specific heat x temp change

we know that

specific heat of water = 4.184 J / g / C

so

Heat = m x s x dT

so

Heat = 115 x 4.184 x ( 95-39)

Heat = 26944.96 J

so

the heat needed is 26945 J

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