Saved Bookmarks
| 1. |
A peiece of metal weight 45g in air and 25g 30^(@)C. When the temperature of the liquide raised to 40^(@) |
|
Answer» SOLUTION :`M_(1) =` mass of the METAL PIECE in air `M_(2) =` mass of the metal piece in LIQUID at `30^(@)C` `M_(3) =` Mass of the metal piece in liequid at `40^(@)C` `rho_(1) =` density of liquid at `30^(@)C` `rho_(2) =` density of liquid at `40^(@)C` Mass expelled `Deltam_(1) = M_(1) - M_(2) = V_(1)rho_(1)` at `30^(@)C` `Deltam_(2) = M_(1) - M_(3) = V_(2)rho_(2)` at `40^(0)C, (20)/(18) = (V_(1)rho_(1))/(V_(2)rho_(2))` `(10)/(9) = (V_(1) xx 1.5 xx 10^(3))/(V_(1)(1 + gammaDeltat) xx 1.25 xx 10^(3)), 1 + gammaDeltat = (9 xx 1.2)/(10)` `= (10.8)/(10) = 1.08, gammaDeltat = 1.08 - 1, 3alpha = 0.08` `ALPHA = (0.08)/(3) = :. alpha = 2.6 xx 10^(-3//@)C` |
|