Question

The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm pressure. 1. q = 6020 j/mol

Answer 1

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$  (in terms of mass)

or, $q=\Delta H_{fusion}\times Moles$     (in terms of moles)

Now we have to calculate the value of q.

$q=6020J/mole\times 1Mole=6020J$

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of $\Delta e=0$

Now we have to calculate the value of w.

Formula used :    $\Delta e=q+w$

where, q is heat required, w is work done and $\Delta e$ is internal energy.

Now put all the given values in above formula, we get

$0=6020J+w$

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0