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A `10 kW` drilling machine is used to drill a bore in a small aluminium block of mass `8.0 kg`. How 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 `= 0.91 J//g^(@)C`. |
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Answer» Given, Power P = 10kW = `10 xx 10^(3)`W Time, t = 2.5min = `2.5 xx 60s` As power is defined as rate at which energy is consumed, i.e. `P=E/t)` `therefore` Total energy used E = `Pt xx 10^(3) xx 2.5 xx 60` = `1.5 xx10^(6)`J Energy absorbed by aluminium block, Q = 50% of the total energy = 50% pf `1.5 xx 10^(6)` = `0.75 xx 10^(6)`J Also, m = 8.0Kg = `8.0 xx 10^(3)`g, c = 0.91 `Jg^(-1^(@)C^(-1)` As, `Q = mc DeltaT` `therefore` `DeltaT = Q/(mc)=(0.75xx10^(6)(8.0 xx10^(3)xx 0.91)` = `(103.02^(@)C` |
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