Solution:

Given:

Power = 10 kW

Mass of the small aluminium block, m = 8 kg = 8 x 10³ g

Time = 2.5×60=150 s

Specific heat of aluminium, c= 0.91 J g^–1 K^–1.

Total energy =P×t=10⁴ ×150=15×10⁵ J

As 50% of the energy is used in the heating or lost to the surrounding,

Therefore, thermal energy available, ΔQ= (1/2) x 15×10⁵
= 7.5 x 10⁵

As we know that, ΔQ=mcΔT
Thus, ΔT= ΔQ/mc

ΔT=  7.5 x 10⁵/ (8 x 10³ ×0.91)

ΔT= 7.5 x 10⁵/ 7.28 x 10³

ΔT= 1.03 x 10² 

Answer: Rise in the temperature of the block, ΔT = 103⁰ C