Question

How much heat (in kJ) is needed to convert 866 g of ice at -10˚C to steam at 126 ˚C? (The specific heats of ice and steam are 2.03 J/g. ˚C and 1.99 J/g. ˚C, respectively.)

Answer 1

2714kJ

Explanation:

This'll be a bit lengthy so bear with me.

This problem will require 5 values;

The heat required to raise the ice to its melting point (0°C), `q_1`

The heat required to melt all of said ice, `q_2`

The heat required to raise the temperature of the water to its boiling point (100°C), `q_3`

The heat required to vaporize all the water, `q_4`

And the heat required to raise the temperature of the steam to 126°C, `q_5`

We can find every value using `q = mcDeltaT`, `q = mH_(fusion)`, and `q = mH_("vaporization")`

`q` = Heat/energy
`m` = Mass
`c` = Specific heat
`DeltaT` = Change in temperature
`H_(fusion)` = Energy required to melt/freeze one gram of a substance
`H_("vaporization")` = Energy required to vaporize/condense one gram of a substance

I'll now write out every equation with the given info.

`q_1 = 866g * (2.03J)/(g°C) * 10°C`

`q_2 = 866g * (334J)/g`

`q_3 = 866g * (4.18J)/(g°C) * 100°C`

`q_4 = 866g * (2260J)/g`

`q_5 = 866g * (1.99J)/(g°C) * 26°C`

Note: in the 3rd and 4 equations, 334J/g and 2260J/g are the heat of fusion and vaporization of water respectively.

After calculating and using sig figs in each equation, we are given 5 values:

`q_1 = 20000J`
`q_2 = 289000J`
`q_3 = 400000J`
`q_4 = 1960000J`
`q_5 = 45000J`

After adding all the values, we have a total of 2714000J, or 2714kJ required for the conversion.