Specific/Latent Heat and Melting Iron

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To calculate the heat required to melt 16.7 kg of iron at 21.3°C, the initial temperature change (Q1) is calculated using the specific heat capacity, yielding 1.14e7 J. The latent heat (Q2) is determined using varying values for iron's latent heat, leading to a total heat requirement of approximately 1.55x10^7 J. There is confusion regarding the units of specific heat capacity, as it should be in kJ/kg*K, which affects the calculations. The discrepancies in the latent heat values for different types of iron also contribute to the uncertainty in the final answer. Clarification on the correct latent heat value and unit consistency is essential for accurate calculations.
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Homework Statement


How much heat is needed to melt 16.7 kg of iron that is initially at 21.3°C?

Homework Equations


Q=m*c*ΔT + mL

The Attempt at a Solution


here: m=16.7kg, c=0.45kJ/kg*K, and ΔT=1538-21.3=1516.7

Thus, Q1=1.14e7 J

Then, I need to calculate latent heat needed, which is Q2 = (2.67 x 10^5)* (16.7) = 0.44584 x 10^7

So, the total heat needed is: Q1+Q2=1.55x10^7 J. I also got 1.58x10^7 on my second try.

But neither are the right answer. What did I do wrong?
 
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Anony-mouse said:

Homework Statement


How much heat is needed to melt 16.7 kg of iron that is initially at 21.3°C?

Homework Equations


Q=m*c*ΔT + mL

The Attempt at a Solution


here: m=16.7kg, c=0.45kJ/kg*K, and ΔT=1538-21.3=1516.7

Thus, Q1=1.14e7 J

Then, I need to calculate latent heat needed, which is Q2 = (2.67 x 10^5)* (16.7) = 0.44584 x 10^7

So, the total heat needed is: Q1+Q2=1.55x10^7 J. I also got 1.58x10^7 on my second try.

But neither are the right answer. What did I do wrong?

I think Q1 is 1.14e4 J (according to my calculator).

But I found like 3 values for the latent heat of iron

Iron, gray cast 96 kJ/kg
Iron, white cast 138 kJ/kg
Iron, slag 209 kJ/kg


So not too sure which one is correct for the question.
 
rock.freak667 said:
I think Q1 is 1.14e4 J (according to my calculator).

J or kJ?
 
Borek said:
J or kJ?

ah didn't notice the units of c...but I'd still get the OP's answer for the 2nd try using that value of the latent heat.
 
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