Saved Bookmarks
| 1. |
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium =0.91" Jg"^(-1)K^(-1). |
|
Answer» Solution :If the consumed energy to DRILL is Q, then `Q=Pt` `=10^(4)xx2.5xx50` `=15xx10^(5)J` Energy OBTAINED by Aluminium block is `Q.=50%` of Q `:.Q.=(15xx10^(5)xx50)/(100)` `:.Q.=75xx10^(4)J` If increase in TEMPERATURE of block is `Deltatheta` then, `Q.="MC"Deltatheta` `:.Deltatheta=(Q.)/("mc")` `=(75xx10^(4))/(8xx0.91xx10^(3))` `=103.2^(@)C` `~~103^(@)C` |
|