Question

Given a consistent source of heat, about how much longer would it take to boil 100 g of water at 100${}^\circ $C than to melt 100 g of ice at 0${}^\circ $C?

Answer 1

Hint: Enthalpy or the value of energy needed for different phases by 1 g of water, that is fusion and vaporization enthalpies are; enthalpy of fusion, $\Delta {{H}_{fus}}$= 333.55 J/g, and enthalpy of vaporization, $\Delta {{H}_{vap}}$= 44.660 kJ/g.

Formula used: power formula (to obtain time), P = $\dfrac{work\,done}{time\,taken}$

Complete step by step answer:
We are given 100 g of sample water and ice, which is observed in two cases. The first case involves boiling of water at 100${}^\circ $C. The second case involves the melting of ice at 0${}^\circ $C. These cases are governed by a consistent source of heat.
We have to find the time taken in both the cases. For this we have to calculate, first the energy for 100 g of samples, and through that energy we will calculate time, using a power formula.
The change in phase when the water is boiled is governed by enthalpy of vaporization, so, for 1 g of water to boil, the value of enthalpy of vaporization, $\Delta {{H}_{vap}}$= 44.660 kJ/g.
So, for 100 g of water to boil, the enthalpy is calculated by conversion or unitary method as:
$\Delta H$For 100 g water = $\dfrac{\Delta {{H}_{vap}}\,of\,1\,g\,water}{1\,g}\times $100 g
$\Delta H$For 100 g water = $\dfrac{44.660\,kJ}{1\,g}\times $100 g
$\Delta H$For 100 g water = 4.4660 kJ = 4466000 J
Similarly, the change in phase when ice is melted is governed by enthalpy of fusion, so, for 1 g of water to boil, the value of enthalpy of fusion, $\Delta {{H}_{fus}}$= 333.55 J/g.
So, for 100 g ice to melt, the enthalpy is calculated by conversion or unitary method as:
$\Delta H$For 100 g ice = $\dfrac{\Delta {{H}_{fus}}\,of\,1\,g\,\,ice}{1\,g}\times $100 g
$\Delta H$For 100 g ice = $\dfrac{333.55\,J}{1\,g}\times $100 g
$\Delta H$For 100 g ice = 33355 J
So, enthalpies of energy for water to boil is 4466000 Joules and for the ice to melt is 33355 Joules.
Now, from the definition of power we have, 1 watt of power consumed by 1 joule energy in 1 second. So, 1 W = $\dfrac{1\,joule}{1\,\sec }$. Assuming that the power is the same for both cases to be 1 watt, now by using a unitary method or conversion factor, we will determine the time taken in the obtained energy for both cases.
Time needed for water to boil = 4466000 J $\times \dfrac{1\,s}{1\,J}$
Time needed for water to boil = 4466000 s
Time needed for ice to melt = 33355 J$\times \dfrac{1\,\sec }{1\,J}$
Time needed for ice to melt = 33355 s
Now clearly, time needed for water to boil is more, so the ratio of both will be, time taken to boil upon time taken to melt,
\[\dfrac{time\,to\,boil}{time\,to\,melt}\]= $\dfrac{4466000\,s}{33355\,\,s}$ = 134 s
Hence, time taken to boil 100 g water is 134 second times more than the time taken to melt.

Note: The given value of enthalpy of vaporization, $\Delta {{H}_{vap}}$= 44.660 kJ/g, is given in kilojoules, that has been converted to joules by the fact that, 1 kJ = ${{10}^{3}}$ J.