PHY..............

Question

A heater is used to heat 3 kg of aluminium for 4 minutes . The rise in the temperature of aluminium is 30'C . If only 80% of the energy supplied by the heater is transferred to aluminium , find the power of the heater . The specific heat capacity of aluminium is 900J kg-1 'C-1 .

Answer 1

E = mcΔT
E = (3)(900)(30)
E = 81000J

the energy supplied = 81000÷0.8 = 101250J

P = E / t
P = (101250) / (4 × 60)
P = 422W

Answer 2

Let y (W) be the power of the heater.

Energy absorbed by aluminium = mcΔT = 3 x 900 x 30 = 81000 J
Energy supplied by heater = Pt = y (4 x 60) = 240y J

(Energy supplied by heater) x 80%= Energy absorbed by aluminium
(240y) x 80% = 81000
Hence, y =421.9 (W)
The power of the heater = 421.9 W