Power of the drilling machine, P = 10 `kW = 10 × 10^(3) W`
Mass of the aluminum block, m = 8.0 `kg = 8 × 10^(3) g`
Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s
Specific heat of aluminium, `c = 0.91 J g^(–1) K^(–1)`
Rise in the temperature of the block after drilling = δT
Total energy of the drilling machine = Pt
`= 10 × 10^(3) × 150`
` = 1.5 × 10^(6) J`
It is given that only 50% of the power is useful.
Useful energy, `DeltaQ=(50)/(100)xx1.5xx10^(6)=7.5xx10^(5)J`
But `DeltaQ=mcDeltaT`
`:.DeltaT=(DeltaQ)/(mc) `
`=(7.5xx10^(5))/(8xx10^(3)xx0.91)`
`=103^(@)C`
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is `103^(@)C`.
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`.
Hira Kakar
Asked: 2 years ago2022-11-03T01:14:49+05:30
2022-11-03T01:14:49+05:30In: General Awareness
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`.
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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Tushar Jhaveri
Asked: 2 years ago2022-10-29T17:12:15+05:30
2022-10-29T17:12:15+05:30In: General Awareness
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. Find the rise in temperature of the block in `2.5` minutes, assuming `50%` power is used up in heating the machine itself or lost to the surropundings.
(Specific heat of aluminium `= 0.91 J g^(-1).^(@)C^(-1)`)
A. `100 .^(@)C`
B. `103 .^(@)C`
C. `150 .^(@)C`
D. `155 .^(@)C`
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. Find the rise in temperature of the block in `2.5` minutes, assuming `50%` power is used up in heating the machine itself or lost to the surropundings.
(Specific heat of aluminium `= 0.91 J g^(-1).^(@)C^(-1)`)
A. `100 .^(@)C`
B. `103 .^(@)C`
C. `150 .^(@)C`
D. `155 .^(@)C`
(Specific heat of aluminium `= 0.91 J g^(-1).^(@)C^(-1)`)
A. `100 .^(@)C`
B. `103 .^(@)C`
C. `150 .^(@)C`
D. `155 .^(@)C`
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Leave an answer
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Correct Answer – B
Here, `P = 10 kW = 10^(4) W, m = 8 kg`
time, `t = 2.5` minute `= 2.5 xx 60 = 150 s`
Specific heat, `s = 0.91 j g^(-1) C^(-1)`
Total energy `= P xx t = 10^(4) xx 150 = 15 xx 10^(5) J`
As `50%` of energy is lost,
`:.` Energy available, `DeltaQ = 1/2 xx 15 xx 10^(5) = 7.5 xx 10^(5) J`
As `DeltaQ = msDeltaT`
`DeltaT = (DeltaQ)/(ms) = (7.5 xx 10^(5))/(8 xx 10^(3) xx 0.91) ~~ 103 .^(@)C`
Here `P = 10 kW =10^4 W, mass m = 8.0 kg = 8 xx10^3 g`
rise in temp. `Delta = ?, time , t = 2.5 min = 2.5 xx 60 = 150 s`
Sp. Heat, `s = 0.91 J g^(-1) .^@C^(-1)`
Total energy, ` = P xx t = 10^4xx150 = 15xx10^5 J`
As, 50 % of energy is lost, `:.` Energy available, `Delta Q = (1)/(2)xx15 xx 10^5 = 7.5 xx 10^5 J`
As `Delta Q =ms Delta T :. Delta T = (Delta Q)/(ms) = (7.5 xx10^5)/(8xx10^3 xx 0.91) = 103 .^@C`