HOTDOG #12 part B

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805132275
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Joined: Fri Sep 28, 2018 12:26 am

HOTDOG #12 part B

Postby 805132275 » Tue Feb 12, 2019 8:53 pm

I don't understand the values for the temperature change used for the first enthalpy changes where glucose and oxygenate added together? Where did 11.3 for glucose and 16.3 for oxygen come from?

Becky Belisle 1A
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Re: HOTDOG #12 part B

Postby Becky Belisle 1A » Tue Feb 12, 2019 9:26 pm

The problem provides the change in enthalpy at 200°C, so the first step is to heat up the reactants from 37°C to 200°C. This is why the change in temperature equals 163°C. I do not see any values of 11.3 or 16.3.

805132275
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Joined: Fri Sep 28, 2018 12:26 am

Re: HOTDOG #12 part B

Postby 805132275 » Tue Feb 12, 2019 9:43 pm

I keep plugging in (1)(219.2)(163)+(6)(29.4)(163) but not getting the correct answer?

Becky Belisle 1A
Posts: 81
Joined: Fri Sep 28, 2018 12:18 am

Re: HOTDOG #12 part B

Postby Becky Belisle 1A » Tue Feb 12, 2019 10:04 pm

When I use the equation you provided, I get 64,482.8 J, which equals the answer provided in the review session.

Kevin Tang 4L
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Re: HOTDOG #12 part B

Postby Kevin Tang 4L » Tue Feb 12, 2019 11:18 pm

(1)(219.2)(163)+(6)(29.4)(163) only gives you part of the answer as you need to heat it up from 37 to 200, then use the DeltaHrxn then cool it down from 200 to 37.

DeltaH1=(1)(219.2)(163)+(6)(29.4)(163)
DeltaH2=-2756kJ
DeltaH3=(6)(37.1)(-163)+(6)(75.3)(-163)

DeltaH1+DeltaH2+DeltaH3 Should give you the answer -2801kJ ->2.8*10^3 kJ (sig figs)

Aurbal Popal
Posts: 63
Joined: Fri Sep 28, 2018 12:27 am

Re: HOTDOG #12 part B

Postby Aurbal Popal » Wed Feb 13, 2019 12:21 am

Kevin Tang 4L wrote:(1)(219.2)(163)+(6)(29.4)(163) only gives you part of the answer as you need to heat it up from 37 to 200, then use the DeltaHrxn then cool it down from 200 to 37.

DeltaH1=(1)(219.2)(163)+(6)(29.4)(163)
DeltaH2=-2756kJ
DeltaH3=(6)(37.1)(-163)+(6)(75.3)(-163)

DeltaH1+DeltaH2+DeltaH3 Should give you the answer -2801kJ ->2.8*10^3 kJ (sig figs)


Why is the temperature for delta h3 negative? Since delta h1 is the enthalpy of the reactants, why isn't delta h1 negative?

Gary Qiao 1D
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

Re: HOTDOG #12 part B

Postby Gary Qiao 1D » Wed Feb 13, 2019 4:35 am

Aurbal Popal wrote:
Kevin Tang 4L wrote:(1)(219.2)(163)+(6)(29.4)(163) only gives you part of the answer as you need to heat it up from 37 to 200, then use the DeltaHrxn then cool it down from 200 to 37.

DeltaH1=(1)(219.2)(163)+(6)(29.4)(163)
DeltaH2=-2756kJ
DeltaH3=(6)(37.1)(-163)+(6)(75.3)(-163)

DeltaH1+DeltaH2+DeltaH3 Should give you the answer -2801kJ ->2.8*10^3 kJ (sig figs)


Why is the temperature for delta h3 negative? Since delta h1 is the enthalpy of the reactants, why isn't delta h1 negative?


The reaction first goes from body temp (37C) to 200C for delta h1, meaning Tf-Ti would be 163. Delta h2 is at 200C. Delta h3 would then have to come back down to body temp, meaning the change in T would be 37C-200C = -163.


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