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.)

1 Answer
May 15, 2018

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.