$1300 J$ of heat energy is supplied to raise the temperature of $0.5 k g$ of lead from $20^{o} C$ to $40^{o} C$ . Calculate the specific heat capacity of lead

130 J k g^{- 1} K^{- 1}

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260 J k g^{- 1} K^{- 1}

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650 J k g^{- 1} K^{- 1}

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None of the above

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Solution

The correct option is A $130 J k g^{- 1} K^{- 1}$
$Q = 1300 = m c Δ T = 0.5 \times (40 - 20) \times c$
So, $c = 130 J / k g K$

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Similar questions

1300 J

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1300J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20oC to 40oC. Calculate the specific heat capacity of lead

The correct option is A 130Jkg−1K−1Q=1300=mcΔT=0.5×(40−20)×cSo, c=130J/kg K

$1300 J$ of heat energy is supplied to raise the temperature of $0.5 k g$ of lead from $20^{o} C$ to $40^{o} C$ . Calculate the specific heat capacity of lead

The correct option is A $130 J k g^{- 1} K^{- 1}$
$Q = 1300 = m c Δ T = 0.5 \times (40 - 20) \times c$
So, $c = 130 J / k g K$