New here Need help with Thermodynamics question

In summary: I'm doing this at work soo I'm not fully focused but yes furnaces do have 6 sides lol whoops =Pyour probably right in your values, lol these stupid books are full of mistakes!ok so let's do that again:a = 2(3x2) + 2(2x2) + 2(3x2)a = 32m²300 x 32 = 9600Wso for the source supply do I need to use the specific heat capacity of the furnace and work out quantity of heat transfered?Hmm. Not quite.Think of the heat supplied to the furnace as a source of heat flowing into
  • #1
candii
8
0

Homework Statement



An electric furnace has dimensions 2m x 2m x 3m. When it reaches forging temperature of 1200°C, a 2t block of steel at 20°C is placed in the furnace. If there is a heat loss of 300W/m² from all surfaces, and a constant energy input of 20kW, calculate how long it will take before the steel can be forged.

Homework Equations



Sensible Heat:

Q = mcΔT = mc(T2 - T1)

Q = quantity of heat transferred (J)
m = mass of the material (kg)
c = specific heat capacity (J/kgK)
T1 = initial temperature (°C)
T2 = final temperature (°C)
ΔT = temperature difference (°C) = T2-T1

Latent heat:

Q = mL

L = Latent heat



The Attempt at a Solution



Not quite sure how to attempt it >___< could someone please guide me in solving this?

Thank you for your help!

Candii ^__^
 
Last edited:
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  • #2
candii said:

Homework Statement



An electric furnace has dimensions 2m x 2m x 3m. When it reaches forging temperature of 1200°C, a 2t block of steel at 20°C is placed in the furnace. If there is a heat loss of 300W/m² from all surfaces, and a constant energy input of 20kW, calculate how long it will take before the steel can be forged.

Homework Equations



Sensible Heat:

Q = mcΔT = mc(T2 - T1)

Q = quantity of heat transferred (J)
m = mass of the material (kg)
c = specific heat capacity (J/kgK)
T1 = initial temperature (°C)
T2 = final temperature (°C)
ΔT = temperature difference (°C) = T2-T1

Latent heat:

Q = mL

L = Latent heat



The Attempt at a Solution



Not quite sure how to attempt it >___< could someone please guide me in solving this?

Thank you for your help!

Candii ^__^

Hi Candii, Welcome to Physics Forums.

You'll have to make some sort of attempt to solve the problem before we can see how to help you.

Why don't you start by determining how much heat energy, Q, is required to bring the steel up to forging temperature?
 
  • #3
gneill said:
Hi Candii, Welcome to Physics Forums.

You'll have to make some sort of attempt to solve the problem before we can see how to help you.

Why don't you start by determining how much heat energy, Q, is required to bring the steel up to forging temperature?

Hi thank you for your respone! HaHa cool, I feel like I'm in school again =P

am I on the right track?:

Q = mc(T2 - T1)
Q = 2000 x 0.46(1200 - 20)
Q = 1085.6 MJ
 
  • #4
candii said:
Hi thank you for your respone! HaHa cool, I feel like I'm in school again =P

am I on the right track?:

Q = mc(T2 - T1)
Q = 2000 x 0.46(1200 - 20)
Q = 1085.6 MJ

The method is good, and the result is correct if the metal is cast iron (specific heat 0.46 kJ/kg-°C). Steel generally has a slightly higher specific heat. Were you given the value to use, or are you expected to look up a suitable value for steel?

The next step will be to determine the rate that heat energy flows into the metal from its environment. You have a source of heat and some stated losses. What's left after the losses are subtracted?
 
  • #5
gneill said:
The method is good, and the result is correct if the metal is cast iron (specific heat 0.46 kJ/kg-°C). Steel generally has a slightly higher specific heat. Were you given the value to use, or are you expected to look up a suitable value for steel?

The next step will be to determine the rate that heat energy flows into the metal from its environment. You have a source of heat and some stated losses. What's left after the losses are subtracted?

In my book it states in the Appendix the following:
Cast Iron = 0.42kJ/kgK
Steel = 0.46kJ/kgK

I just followed what was in there =)

ok, let me see so the environment is the furnace itself and the losses stated is 300W/m²
so I should work out the area of the furnace and times that buy the energy loss?

a = l x w
2 x 2
a = 4m²

4 x 300 = 1200w

now kinda stuck =/ should I work out the volume?
 
  • #6
candii said:
In my book it states in the Appendix the following:
Cast Iron = 0.42kJ/kgK
Steel = 0.46kJ/kgK

I just followed what was in there =)
Okay. Follow the book's values to get the book's results :wink:

I've seen published values of 0.46 for cast iron and 0.49 for carbon steel. Take the book value of 0.46.
ok, let me see so the environment is the furnace itself and the losses stated is 300W/m²
so I should work out the area of the furnace and times that buy the energy loss?
Yes.
a = l x w
2 x 2
a = 4m²

4 x 300 = 1200w
Umm. The furnace has six sides...
now kinda stuck =/ should I work out the volume?
No, the volume is not important. You need to find the total heat loss rate for the surface area of the furnace. The heat supplied to the metal will be the source supply of heat less the losses.
 
  • #7
gneill said:
Okay. Follow the book's values to get the book's results :wink:

I've seen published values of 0.46 for cast iron and 0.49 for carbon steel. Take the book value of 0.46.

Yes.

Umm. The furnace has six sides...

No, the volume is not important. You need to find the total heat loss rate for the surface area of the furnace. The heat supplied to the meta will be the source supply of heat less the losses.

lmfao >__< I'm doing this at work soo I'm not fully focused but yes furnaces do have 6 sides lol whoops =P

your probably right in your values, lol these stupid books are full of mistakes!

ok so let's do that again:


a = 2(3x2) + 2(2x2) + 2(3x2)
a = 32m²

300 x 32 = 9600W


so for the source supply do I need to use the specific heat capacity of the furnace and work out quantity of heat transferred like I did with the steel? and then work it out?
 
  • #8
candii said:
ok so let's do that again:


a = 2(3x2) + 2(2x2) + 2(3x2)
a = 32m²

300 x 32 = 9600W
Looks good.
so for the source supply do I need to use the specific heat capacity of the furnace and work out quantity of heat transferred like I did with the steel? and then work it out?

You're given a value for the rate at which heat is supplied to the furnace in watts. The watt is defined to be a joule/second. You can assume that the portion of this heat that does not end up being lost through the walls will end up in the steel. So how many seconds are required to supply enough heat to the steel?
 
  • #9
Is the value the constant energy input of 20kw? So in joules per second is it 20000? Lol I'm pretty stuck here, :/
 
  • #10
candii said:
Is the value the constant energy input of 20kw? So in joules per second is it 20000?
Yes. That's the rate that heat energy is being supplied. You've calculated the rate at which heat is lost by the system, so what's the rate at which heat is going into the steel? How much heat does the steel require?
 
  • #11
Ok so after I subtract the loss of 9600w from the 20000w I get 10400w and the steel needs to reach 1200'C to reach forging temperature
 
  • #12
check the rate of "heat transfer" that you've already calc. at the beginning ;) ...from 20°C up to 1200°C
 
  • #13
candii said:
Ok so after I subtract the loss of 9600w from the 20000w I get 10400w and the steel needs to reach 1200'C to reach forging temperature

So heat is added at the rate of 10400 J/s and it requires Q = 1085.6 MJ ...
 
  • #14
Ohhhhh I get it!
1085600kj/s / 10.4kj/s = 104384.62s / 3600s = 29hrs?
 
  • #15
candii said:
Ohhhhh I get it!
1085600kj/s / 10.4kj/s = 104384.62s / 3600s = 29hrs?

That looks like a reasonable result :smile:
 
  • #16
Ohhhhh haha thanku! I get so lost our teacher is good but he has a really hard way of teaching that everyone in our class has trouble understanding so we have to teach ourselves lol thank you so much :)
 

Related to New here Need help with Thermodynamics question

### 1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the conversion of energy from one form to another and the relationship between heat, work, temperature, and energy. It also studies the behavior of systems at the macroscopic level and the laws that govern their behavior.

### 2. What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but it can be transferred from one form to another or converted from one system to another. This means that the total energy of an isolated system remains constant.

### 3. What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of an isolated system will never decrease over time. Entropy is a measure of the disorder or randomness in a system. This law also states that heat will always flow from a hot object to a cold object, never the other way around, unless external work is done.

### 4. How do you calculate the efficiency of a thermodynamic process?

The efficiency of a thermodynamic process is calculated by dividing the work output by the total energy input. This is known as the Carnot efficiency and is given by the equation: η = (Thot - Tcold) / Thot, where Thot is the temperature of the hot reservoir and Tcold is the temperature of the cold reservoir.

### 5. What is the difference between heat and temperature?

Heat and temperature are related but distinct concepts in thermodynamics. Heat is the transfer of thermal energy from one object to another due to a difference in temperature. Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a substance. In other words, heat is a form of energy, while temperature is a measure of the intensity of heat.

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