How much heat to change 1 mole of ice at -25 °c to steam at +125 °c? • heat to warm ice - heat to melt ice (no temperature change) - heat to warm water • heat to boil water (no temperature change) - heat to warm steam

Answer : The amount of heat changes is, 56.463 KJ

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change or heat changes = ?

n = number of moles of water = 1 mole

= specific heat of solid water =

= specific heat of liquid water =

= specific heat of liquid water =

m = mass of water

= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

    (1 KJ = 1000 J)

Therefore, the amount of heat changes is, 56.463 KJ

2b2t

explanation:

2b2t