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Chapter 8

Chapter 8

Postby Legend786 » Thu May 15, 2014 10:33 pm

Given the data in Chart 1, how much would it cost to maintain the pyramid’s current internal temperature for a full year if no modifications to it were made? (Note: Assume that there are 365 days in a year, that the pyramid will need to be cooled year round, and that 100,00 btus can be purchased for $0.50)

A) $3,416
B) $159
C) $3,832
D) $1,708

Result: Your answer is correct.
Your choice: C Correct choice: C

Explanation: The total amount of energy transfer per day for the pyramid is given by 24 (Qb+4Qi). Qb and Qi are given; QT, the total amount of heat transfer, may be calculated at 2.1 x 106 btu. Heat transfer for the entire year is obtained by multiplying this figure by 365. Subsequently, dividing by 1x 105 yields the number (n) of costs units at $0.50 each. n multiplied by the cost is the correct answer

Tin = 55 oC

Qb = 2.3 x 104 btu

Qi = 1.6 x 104 btu

Rb = 2

Rf = 5

Tout = 105 oF



Where Tin is the temperature inside the pyramid, Qb is the total heat transfer through the base, Qi is the total loss through one side, Rb is the R-value of the material of which the base is composed, Rf is the value of the material of which the sides are composed, and Tout is the temperature outside of the pyramid.



Hi there does anyone know how the value of QT was calculated?....

Thanks
Legend786
 
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Re: Chapter 8

Postby goldstanda3269 » Mon May 19, 2014 3:58 pm

Qb = 2.3 x 10^4

Qi = 1.6 x 10^4


QT = 24 (Qb+4Qi) = 24 (8.7 x 10^4) = 209 x 10^4 = 2.09 x 10^6 = 2.1 x 10^6
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Re: Chapter 8

Postby olirushworth » Mon Jul 07, 2014 8:06 am

Thank you for the answer for this one as it was one I struggled with to figure out how it was worked out.

I followed up with:
2.1 x 10^6 x 365 = 2 x 10^6 = 730 x 10^6 (approx)
730 x 10^6 = 730,000,000
730,000,000/100,000 = 7,300
7,300 x 1/2 = 3750 which is nearer to $3,832

I'm not sure if that is too long winded but seemed the easiest way in my head.
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