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* 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* c_(p,s)* (T_(final)-T_(initial))]+m* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]+m* \Delta H_(vap)+[m* c_(p,g)* (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* 2.09J/g^oC* (0-(-20))^oC]+10.0g* 333J/g+[10.0g* 4.18J/g^oC* (100-0)^oC]+10.0g* 2260J/g+[10.0g* 2.03J/g^oC* (120-100)^oC]`

`\Delta H=30934J=3.09* 10^4J`

Therefore, the amount of heat required is,
`3.09* 10^4J`