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`


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