Question

1.00 kg of ice at -10 °C is heated using a Bunsen burner flame until all the ice melts and the temperature reaches 95 °C. A) How much energy in kJ is required to effect this transformation?

Answer 1

Answer : The energy required is, 574.2055 KJ

Solution :

The conversions involved in this process 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)(95^oC)`

Now we have to calculate the enthalpy change or energy.

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

where,

`\Delta H` = energy required = ?

m = mass of ice = 1 kg = 1000 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`

n = number of moles of ice =
`\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=(1000g)/(18g/mole)=55.55mole`

`\Delta H_(fusion)` = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

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

`\Delta H=[1000g* 4.18J/gK* (0-(-10))^oC]+55.55mole* 6010J/mole+[1000g* 2.09J/gK* (95-0)^oC]`

`\Delta H=574205.5J=574.2055kJ` (1 KJ = 1000 J)

Therefore, the energy required is, 574.2055 KJ