#### 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 = 0.91 J g^{–1} K^{–1}

#### Solution 1

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 = 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, `triangle Q = 50/100 xx 1.5 xx 10^6 = 7.5xx10^5 J`

But `triangle Q = mctriangle T`

`:. triangle T = (triangle Q)/"mc"`

`= (7.5 xx 10^5)/(8xx10^3xx0.91)`

`= 103 ^@C`

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

#### Solution 2

Power = 10 kW = 10^{4} W

Mass, m=8.0 kg = 8 x 10^{3} g

Rise in temperature, `triangle T =?`

`Time, t = 2.5 min = 2.5 xx 60 = 150 s`

Specific heat, `C = 0.91 Jg^(-1) K^(-1)`

Total energy = Power x Time = `10^4 xx 150 J`

`=15 xx 10^5 J`

As 50% of energy is lost

∴Thermal energy available

`triangle Q = 1/2 xx 15 xx 10^5 = 7.5 xx 10^5 J`

Since `triangle Q = mctrangle T`

`:.triangle T = triangleQ/mc = (7.5xx20^5)/(8xx10^3xx0.91) = 103^@C`