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What is the minimum amount of energy required to completely melt a 7.25-kg lead brick which has a starting temperature of 18.0 °C? The melting point of lead is 328 °C. The specific heat capacity of lead is 128 J/(kg∙C°); and its latent heat of fusion is 23,200 J/kg.a. 1.68 × 105 J b. 2.88 × 105 J c. 4.56 × 105 J d. 5.96 × 105 J e. 7.44 × 105 J

Respuesta :

tutorAnne

Answer: c. 4.56 × 105 J

Explanation:

Given that

mass of lead brick, m= 7.25kg

Temperature T1 =  18.0 °C

Temperature T2 = 328 °C

specific heat capacity of lead, c = 128 J/(kg∙C°)

latent heat of fusion Lfusion =23,200 J/kg

Amount of energy Q =?

Using the formulae

Amount of energy ,Q =mc ( T2-T1)+ mLfusion

7.25kg x 128 J/(kg∙C°) x (328-18°C) + 7.25kg x 23200 J/kg

=455880J

=4.56 x 10^5 J

The minimum amount of energy required is 4.56 × 10^5 J.

Calculation of the minimum amount of energy:

Since

mass of lead brick, m= 7.25kg

Temperature T1 =  18.0 °C

Temperature T2 = 328 °C

specific heat capacity of lead, c = 128 J/(kg∙C°)

latent heat of fusion =23,200 J/kg

So

we know that

Amount of energy ,Q =mc ( T2-T1)+ mLfusion

7.25kg x 128 J/(kg∙C°) x (328-18°C) + 7.25kg x 23200 J/kg

=455880J

=4.56 x 10^5 J

hence, The minimum amount of energy required is 4.56 × 10^5 J.

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