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3 years ago

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how much heat is needed to melt 2.5 kg of lead? The latent heat of fusion for lead is 5.85 Kcal/kg. The latent heat of vaporizat

ion of lead is 20.8 kcal/kg

1 answer:

sergey [27]3 years ago

5 0

I think its 550 deggres celsius.

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3 years ago

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3 years ago

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4 years ago

Read 2 more answers

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe

JulijaS [17]

Answer:

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.$

Explanation:

Given charges are:

$\rm q_1 = +6e.\\q_2 = -6e.$

The electric potential energy of a charge due to the electric field of another charge is given by

$\rm EPE=\dfrac{kq_1q_2}{r}.$

where,

k = Coulomb's constant, having value = $\rm 9\times 10^9\ Nm^2/C^2.$
r = distance between the charges.

When the charges are infinite distance apart, $\rm r = \infty$ ,

$\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.$

When the charges are $\rm 5.61\times 10^{-12}\ m$ apart, $\rm r=5.61\times 10^{-12}\ m$ ,

$\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.$

Here, e is the charge on one electron, such that, $\rm e = -1.6\times 10^{-19}\ C$ .

Therefore,

$\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.$

Thus,

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.$

4 0

3 years ago