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 surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`.

Explanation

Power of the drilling machine, P = 10 `kW = 10 × 10^(3) W`
Mass of the aluminum block, m = 8.0 `kg = 8 × 10^(3) g`
Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s
Specific heat of aluminium, `c = 0.91 J g^(–1) K^(–1)`
Rise in the temperature of the block after drilling = δT
Total energy of the drilling machine = Pt
`= 10 × 10^(3) × 150`
` = 1.5 × 10^(6) J`
It is given that only 50% of the power is useful.
Useful energy, `DeltaQ=(50)/(100)xx1.5xx10^(6)=7.5xx10^(5)J`
But `DeltaQ=mcDeltaT`
`:.DeltaT=(DeltaQ)/(mc) `
`=(7.5xx10^(5))/(8xx10^(3)xx0.91)`
`=103^(@)C`
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is `103^(@)C`.