Question

Calculate the amount of heat required to convert 10.0 grams of ice at –20.°C to steam at 120.°C. (Sp. heat of H2O(s) = 2.09 J/g•°C, Sp. heat of H2O(l) = 4.18 J/g•°C, Sp heat of H2O(g) = 2.03 J/g•°C; heat of fus. of H2O(s) = 333 J/g, heat of vap. of H2O(l) = 2260 J/g).

Answer 1

Answer : The amount of heat required is, $3.09\times 10^4J$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(-20^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(120^oC)$

The expression used will be:

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

where,

$\Delta H$ = heat required for the reaction

m = mass of ice = 10.0 g

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

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

$c_{p,g}$ = specific heat of gaseous water = $2.03J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $333J/g$

$\Delta H_{vap}$ = enthalpy change for vaporization = $2260J/g$

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

$\Delta H=[10.0g\times 2.09J/g^oC\times (0-(-20))^oC]+10.0g\times 333J/g+[10.0g\times 4.18J/g^oC\times (100-0)^oC]+10.0g\times 2260J/g+[10.0g\times 2.03J/g^oC\times (120-100)^oC]$

$\Delta H=30934J=3.09\times 10^4J$

Therefore, the amount of heat required is, $3.09\times 10^4J$