A) \[10.23\text{ }{}^\circ C\]
B) \[23.10{}^\circ C\]
C) \[103.02\text{ }{}^\circ C\]
D) \[301.02{}^\circ C\]
Correct Answer: C
Solution :
| total energy used by drilling machine \[=Pt\]\[=\left( 10\times {{10}^{3}} \right)\times \left( 2.5\times 60 \right)\]\[=1.5\times {{10}^{6}}\operatorname{J}\] |
| The energy absorbed by the aluminium block \[=\frac{50}{100}\times 1.5\times {{10}^{6}}\] |
| Let \[\Delta T\]be the rise in temperature of the aluminium block, then |
| \[mC\Delta T=0.75\times {{10}^{6}}\]\[\left( 8\times {{10}^{3}} \right)\times 0.91\times \Delta T=0.75\times {{10}^{6}}\] |
| \[\therefore \Delta T=103.02{}^\circ C\] |
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