__Question 11.12__:
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 J g

^{–1}K^{–1}.

__Solution__:*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 = δ

*T*

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, ∆Q = (50/100) X 1.5 X 10

^{6}= 7.5 X 10

^{5}J

But ∆Q = mc∆T

∴ ∆T = ∆Q / mc

= (7.5 X 10

^{5}) / (8 X 10

^{3}X 0.91)

= 103

^{o}C

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

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