Question

Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from –10 °C to 182°C. Assume that the specific heat of water is 4.184 J/g · °C for liquid and that the specific heat of steam is 1.99 J/g · °C.

Answer 1

Answer : The amount of heat required is, $1.16\times 10^6J$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(182^oC)$

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 = 346 g

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

$c_{p,l}$ = specific heat of liquid water = $4.184J/g^oC$

$c_{p,g}$ = specific heat of gaseous water = $1.99J/g^oC$

$\Delta H_{fusion}$ = 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:

$\Delta H=[346g\times 2.09J/g^oC\times (0-(-80))^oC]+346g\times 333J/g+[346g\times 4.184J/g^oC\times (100-0)^oC]+346g\times 2260J/g+[346g\times 1.99J/g^oC\times (180-100)^oC]$

$\Delta H=1.16\times 10^6J$

Therefore, the amount of heat required is, $1.16\times 10^6J$