Question

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).

Answer 1

Answer:

$T_2=571.9K$

Explanation:

Hello,

In this case, we consider the following formula defining the energy and the temperature change for the sample of iron:

$Q=n_{Fe}Cp_{Fe}(T_2-T_1)$

Now, solving the final temperature, considering a positive inlet heat, we have:

$T_2=T_1+\frac{Q}{n_{Fe}Cp{Fe}} =293K+\frac{3500J}{0.5mol*25.1J/(mol*K)} \\T_2=571.9K$

Best regards.

Answer 2

Answer:

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