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
Given: Power of the drilling machine is 10 kW , mass of the aluminum block is 8 kg , time for which the machine used is 2.5 min and specific heat of aluminum is 0.91 Jg -1 K -1 .
The total energy of the drilling machine is given as,
E = P × t
Where, the power of the drilling machine is P and the time for which the machine used is t .
By subtitling the given values in the above formula, we get
E = 10 × 10 3 × 2.5 × 60 = 1.5 × 10 6 J
It is given that only 50 % of the power is useful. So, the useful energy is given as,
Δ Q = 1.5 × 10 6 × ( 50 100 ) = 7.5 × 10 5 J
The rise in temperature is given as,
Δ Q = m c Δ T Δ T = Δ Q m c
Where, mass of the substance is m , specific heat is c and the change in temperature is Δ T .
By substituting the given values in the above expression, we get
Δ T = 7.5 × 10 5 ( 8 × 10 3 ) × 0.91 ≈ 103 ° C
Thus, in 2.5 min of drilling, the rise in temperature of the block is 103 ° C .
Given: Power of the drilling machine is
The total energy of the drilling machine is given as,
Where, the power of the drilling machine is
By subtitling the given values in the above formula, we get
It is given that only
The rise in temperature is given as,
Where, mass of the substance is
By substituting the given values in the above expression, we get
Thus, in
