Verified answer

This problem must be worked in stages.

Stage 1: 125 g Steam @ 110.°C --> 125 g steam @ 100.°C Use H(1) = mcΔT

m = 125 g; c = heat capacity of steam = 1.86 J/g*°C; ΔT = (100.°C - 110°C)

Stage 2: 125 g steam @ 100.°C --> 125 g H2O(l) @ 100.°C Use H(2) = mLv

m = 125 g; Lv = 40.7 J/g

Stage 3: 125 g H2O(l) @ 100.°C --> 125 g H2O(l) @ 0.°C Use H(3) = mcΔT

c = 4.184 J/g*°C; ΔT = (0.°C - 100.°C)

Stage 4: 125 g H2O(l) @ 0.°C --> 125 g H2O(s) @ 0.°C Use H(4) = mLf

Lf = 6.01 J/g*°C

Stage 5: 125 g H2O(s) @ 0.°C --> 125 g H2O(s) @ -5.00°C Use H(5) = mcΔT

c = 2.06 J/g*°C; ΔT = (0.°C - 5.00°C)

Total heat removed = H(1) + H(2) + H(3) + H(4) + H(5)

Now calculate each one and add them up. Try to include the units so that you will know what you have when you are finished.

H(1) = mcΔT = (125 g)(1.86 J/g*°C)(100.°C - 110°C) = -2325 J

H(2) = mLv = (125 g)(40.7 J/g) = -5088 J

H(3) = (125 g)(4.184 J/g*°C)(0.°C - 100.°C) = -52300 J

H(4) = (125 g)(6.01 J/g) = -751 J

H(5) = (125 g)(2.06 J/g*°C)(-5.00°C) = 1288 J

Now add them together and you will have your answer. The negative sign tells you that energy is removed.

Hope this is helpful to you. JIL HIR

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