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)`.

Explanation

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`