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 surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`.

Explanation

P= 10 kW, t = 2.5 min `= 2.5 xx 60 s`.
Total energy `= Pxx t = 10^4 xx 2.5 xx 60 J = 15 xx 10^5 J`
Specific heat, `C = 0.91 Jg^(-1)K^(-1)`
` = 0.91 xx 10^3 J kg^(-1) K^(-1)`
As 50% energy is used up in raising the temperature, thermal energy available
`Delta Q = 1/2 xx 15 xx 10^5 J = 7.5 xx 10^5 J`
As `Delta Q = M xx C xx Delta T`
`M xx C xx Delta T = 7.5 xx 10^5`
`8 xx 0.91 xx 10^3 xx Delta T = 7.5 xx 10^5 `
` Delta T = Delta Q//MC`
` = (7.5 xx 10^5)/(8 xx 0.91 xx 10^3) = 103^@C`