3500 J of energy are added to a 0.5 mol sample of iron at 293 K. What is the final temperature of the iron in kelvins? The molar heat of iron is 25.1 J/(mol•K).

Explanation:

Step 1: Data given

Number of heat transfer = 3500 J

Number of moles of iron = 0.5 moles

Initial temperature = 293 K

The molar heat of iron is 25.1 J/(mol*K)

Step 2: Calculate ΔT

Q = n* C * ΔT

⇒with Q = the heat transfer = 3500 J of energy

⇒with n = the number of moles iron  = 0.5 moles

⇒with C = the molar heat of iron = 25.1 J/mol*K

⇒ΔT = the change of temperature = T2 - T1 = T2 - 293 K

3500 J = 0.5 moles *25.1 J/mol * K * ΔT

ΔT = 278.9

Step 3: Calculate ΔT

ΔT = 278.9 = T2 - T1 = T2 - 293 K

T2 = 278.9 + 293 K

T2 = 551.9 K

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each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i