Question

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 = 0.91J/$1g^{-1} K^{-1}$

Answer 1

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Answer 2

Answer:

• $103^{\circ}C$

Explanation:

• Power of the drilling machine, $P = 10 kW = 10 \times 10^3 W$
• Mass of the aluminium block, $m = 8.0 kg = 8 \times 10^3 g$
• Time for which the machine is used, $t = 2.5 min = 2.5 \times 60 = 150 s$
• Specific heat of aluminium, $c = 0.91 Jg^{-1}K^{-1}$
• The Rise in the temperature of the block after drilling = $\delta T$
• The total energy of the drilling machine = Pt

$= 10\times 10^{3} \times 150$

$= 1.5 \times 10^{6}J$

It is given that only 50% of the power is useful.

Hence, Useful energy, $\Delta Q = \frac{50}{100}\times 1.5\times 10^{6} = 7.5 \times 10^5 J$

But $\Delta Q = mc \Delta T$

$\therefore \: \Delta T = \frac{\Delta Q}{mc}$

$=\frac{7.5 \times 10^5}{8\times 10^3 \times 0.91}$

$= 103^{\circ}C$