Calculate the amount of heat required to melt an ice cube with a mass of 250 g and a temperature of -20 degrees celsius and turn it to steam at 120 degrees celsius

Question

Use joules please. not mole .

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Answer 1

To answer this problem you will need several values which you can look up. The values I use may be slightly different from ones you may have. Here are the values I will be using:

Cice = 2.09 J/gº

Cliquid = 4.184 J/gº

Cgas = 1.89 J/gº

∆Hfus = 334 J/g

∆Hvap = 2260 J/g

We will also be using these equations:

q = mC∆T

q = m∆Hfus

q = m∆Hvap

q = heat; m = mass; C = specific heat; ∆T = change in temperature; ∆Hfus = enthalpy of fusion; ∆Hvap = enthalpy of vaporization

These problems are best done using a stepwise approach.

(1) find heat to raise the temperature of the ice from -20º to 0º

(2) find the heat to melt the ice at 0º (phase change only)

(3) find the heat to raise the temperature of liquid from 0º to 100º

(4) find the heat to turn liquid to gas at 100º (phase change only)

(5) find the heat to raise temperature of gas from 100º to 120º

(1) q = mC∆T = (250 g)(2.09 J/gº)(20º) = 10,450 J

(2) q = m∆Hfus = (250 g)(334 J/g) = 83,500 J

(3) q = mC∆T = (250 g)(4.184 J/gº)(100º) = 104,600 J

(4) q = m∆Hvap = (250 g)(2260 J/g) = 565,000 J

(5) q = mC∆T = (250 g)(1.89 J/gº)(20º) = 9450 J

Add up all the joules to get 773,000 J = 773 kJ