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)`.

Explanation

Given, Power P = 10kW = `10 xx 10^(3)`W
Time, t = 2.5min = `2.5 xx 60s`
As power is defined as rate at which energy is consumed,
i.e. `P=E/t)`
`therefore` Total energy used E = `Pt xx 10^(3) xx 2.5 xx 60` = `1.5 xx10^(6)`J
Energy absorbed by aluminium block,
Q = 50% of the total energy = 50% pf `1.5 xx 10^(6)` = `0.75 xx 10^(6)`J
Also, m = 8.0Kg = `8.0 xx 10^(3)`g, c = 0.91 `Jg^(-1^(@)C^(-1)`
As, `Q = mc DeltaT`
`therefore` `DeltaT = Q/(mc)=(0.75xx10^(6)(8.0 xx10^(3)xx 0.91)` = `(103.02^(@)C`