250 grams water. 80 degrees celsius to 120 degress, how many joules needed

Question

First question on a test i had that i havent gotten back yet, wanted to see if i got it right. my answer was 605,860 Joules

Answer 1

Student,

 

I explained the method in your previous posting of this same question.  Please take a look at it, since I will use the same equations again.  It is a matter of just plugging in the numbers.

 

c₁ = specific heat of liquid water = 4186 J /(kg °C)

 

Q₁ = mcΔT₁ = (0.250 kg)[4186 J /(kg °C)](100 - 80)°C = 20,930 J

 

l_v = latent heat of vaporization of water = 2260 J/g

 

Q₂ = ml_v = (250 g)(2260 J/g) = 565,000 J

 

c₃ = specific heat of steam = 1.996 J/(g deg C)

 

Q₃ = mc₃ΔT₃ = (250 g)[1.996 J/(g deg C)](120 - 100)°C = 9980 J

 

Q = Q₁ + Q₂ + Q₃ = (20,930 + 565,000 + 9,980)J = 595,910 J

 

Our answers are very close.  I suspect the difference is due to small differences in tabulated values of specific heats and latent heats.  Which values did you use for the two specific heats and the latent heat of vaporization?