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

Answer 1

Power of the drilling machine, P = 10 kW = 10 × 10³ W

Mass of the aluminum block, m = 8.0 kg = 8 × 10³ 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 = δT

Total energy of the drilling machine = Pt

= 10 × 10³ × 150

= 1.5 × 10⁶ 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.

Answer 2

Power = 10 kW = 10⁴ W

Mass, m=8.0 kg = 8 x 10³ 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`