Determine the amount of heat gained by 1,250 g of water as it changes from ice at -5 degrees C to steam at 120 degrees C.

Question

I keep getting the wrong answers. It says the correct answer is 3,829,000 J. I just need an explanation. And please don't use mol. Thank you.

Answer 1

For these types of problems, it helps to work in steps, as follows:

Step 1: heat to convert 1250 g ice from -5 to 0 degrees

q = mC∆T

q = (1250 g)(2.09 J/g/deg)(5 deg) = 13,063 J (I used 2.09 as C of ice)

Step 2: heat to melt the ice at 0 degrees - phase change - use ∆Hfusion

q = m∆Hfusion

q = (1250 g)(334 J/g) = 417,500 J

Step 3: heat to raise temp from 0 to 100 degrees

q = mC∆T

q = (1250 g)(4.184 J/g/deg)(100 degrees)

q = 523,000 J

Step 4: heat to turn water to steam at 100 degrees - phase change - use ∆Hvap

q = m∆Hvap

q = (1250 g)(2260 J/g) = 2,825,000 J

Step 5: heat to raise temp of steam from 100 to 120 degrees

q = mC∆T

q = (1250 g)(1.89 J/g/deg)(20 degree) = 47,250 J (I used 1.89 as the C for steam)

Adding up all the joules of heat from the 5 steps, we get...

13,063 + 417,500 + 523,000 + 2,825,000 + 47,250 = 3,395,250 J

This answer is off from your answer by 11% and that may be because of different values of C and/or ∆Hfusion and ∆Hvaporization. But the process shown is correct.