Question

Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, Gee 2.03 Jig C, G team Jig C, molar heat of fusion of ice 6.01 * 10 J/mol; molar heat of vaporization of liquid water 4.07 * 10*J/mol 202 Short Answer Toolbar navigation B I V S E 1 E A A This question will be sent to your instructor for grading

Answer 1

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

`(1):H_2O(s)(-15.5^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)(124^oC)`

Now we have to calculate the enthalpy change.

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

where,

`\Delta H` = enthalpy change or heat required = ?

m = mass of water = 45 g

`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`

n = number of moles of water =
`\frac{\text{Mass of water}}{\text{Molar mass of water}}=(45g)/(18g/mole)=2.5mole`

`\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=[45g* 4.18J/gK* (0-(-15.5))^oC]+2.5mole* 6010J/mole+[45g* 2.09J/gK* (100-0)^oC]+2.5mole* 40670J/mole+[45g* 1.84J/gK* (124-100)^oC]`

`\Delta H=139287.75J=139.28775KJ` (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ