Question

How much heat to change 1 mole of ice at -25 °C to steam at +125 °C? • heat to warm ice - heat to melt ice (no temperature change) - heat to warm water • heat to boil water (no temperature change) - heat to warm steam

Answer 1

Answer : The amount of heat changes is, 56.463 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(125^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat changes = ?

n = number of moles of water = 1 mole

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

m = mass of water

$\text{Mass of water}=\text{Moles of water}\times \text{Molar mass of water}=1mole\times 18g/mole=18g$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

$\Delta H=[18g\times 4.18J/gK\times (0-(-25))^oC]+1mole\times 6010J/mole+[18g\times 2.09J/gK\times (100-0)^oC]+1mole\times 40670J/mole+[18g\times 1.84J/gK\times (125-100)^oC]$

$\Delta H=56463J=56.463KJ$     (1 KJ = 1000 J)

Therefore, the amount of heat changes is, 56.463 KJ