InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Examine table.s and express a) The energy required to break one bond in DNA in eV, b) the kinetic energy of an air molecule `(10^(-2)` J in eV, c) The daily intake of a human adult in kilocalories. |
|
Answer» a) Energy required to break one bond of DNA is: `(10^(-20)/(1.6 xx 10^(-19) J//eV) =~ 0.06 eV` Note, 0.1eV = 100meV (100 millielectron volt). B) the kinetic energy of an air molecule is `(10^(-21))/(1.6 xx 10^(-19)J//.eV=~ 0.0062eV` this is the same as 6.2 meV. c) The average human consumption in a day is `10^(7)/(4.2xx 10^(3)J/kcal) -~2400 kcal` We point out a common misconception created by newspapers and magzines. They mention food values in calories and urge us to restrict diet intake to below 2400 claories. What they should be saying in kilocalories (kcal) and not calories. A person consuming 2400 calories a day will soon starve to death 1 food calorie is 1 kcal. |
|