Correct Answer - B
We are given that power of the machine,
`P=10kW = 10^(4)W`
time for which the machine is used,
`t=2.5min=150s`
Mass of the alunumum block,
`m=8kg = 8xx10^(3)g`
Specific heat of aluminum, `c=0.91J//g^(@)C`
Since `50%` of the energy is used up in heating the machine itself or lost to the surroundings,
Energy absorbed by the block, `Q=0.5Pt`
`=0.5xx10^(4)xx150J = 7.5xx10^(5)J`
Let `Delta T` be the rise in temperature of the block,
As, `Q=mc Delta T`
`Delta T = (Q)/(mc) = (7.5xx10^(5)(J))/(8xx10^(3)(g)xx0.91(J//gC^(@)))=103^(@)C`.