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$\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$.

We are given that power of the machine,

      P = 10 kW = ${10}^4 \mathrm{W}$

Time for which the machine is used,

      t = 2.5 min = 150 s

Mass of the aluminium block,

     m = 8 kg = 8 $\times {10}^3 \mathrm{g}$

Specific heat of aluminium, c = 0.91 $\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Since 50% of the energy is used up in heating the machine itself or lost to the surroundings,

Energy absorbed by the block, Q = 0.5 Pt

    = 0.5 $\times {10}^4\times 150 \mathrm{J} = 7.5 \times {10}^5 \mathrm{J}$

Let $∆\mathrm{T}$ be the rise in temperature of the block.

As Q = mc $∆\mathrm{T} = \frac{\mathrm{Q}}{\mathrm{mc}} = \frac{7.5 \times {10}^5\left(\mathrm{J}\right)}{8\times {10}^3\left(\mathrm{g}\right)\times 0.91(\mathrm{J}/{\mathrm{gC}}^0)} = {103}^0\mathrm{C}$