question_answer 12)

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How 11 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 \[=1\cdot 074\times {{10}^{-2}}m=1\cdot \text{074 cm}\]\[6\cdot 9\times {{10}^{7}}Pa\]

Answer:

Here,\[49\times {{10}^{-5}}\]mass, \[^{\text{o}}{{\text{C}}^{\text{-1}}}\] Rise in temp, \[\gamma =49\times {{10}^{-5}}\] time, \[^{\text{o}}{{\text{C}}^{\text{-1}}}\text{,}\] \[\vartriangle T=30\] Sp. Heat, \[^{o}C\] \[V'=V+\vartriangle V=V\left( 1+\gamma \vartriangle T \right)\] Total energy,= \[\therefore \] As, 50% of energy is lost, \[V'=V\left( 1+49\times {{10}^{-5}}\times 30 \right)=1\cdot 0147V\] Energy available, ` \[\rho =\frac{m}{V},\rho '=\frac{m}{V'}=\frac{m}{1\cdot 0147V}=0\cdot 9855\rho \] As, \[=\frac{\rho -\rho '}{\rho }=\frac{\rho -0\cdot 9855\rho }{\rho }=0\cdot 0145\]= m c\[8\cdot 0\]\[2\cdot 5\] \[=0\cdot 91J{{g}^{-1}}\] \[^{o}{{C}^{-1}}.\]