Specific Heat Capacity of Tantalum

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  • #1
moimoi24
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I got a problem with this one, please help me,

Tantalum is an element that is used in aircraft parts. Tantalum has a specific heat capacity of about 140 J/kg K. The aircraft part has a mass of 0.23 kg and is cooled from a temperature of 1200 K by being placed in water. If 30000 J of heat is transferred to the water, what is the final temperature of the part?

How to solve this one, i hope you can help me. Thanks...
 

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  • #2
CWatters
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Energy = Mass * specific_heat_capacity * Temperature_change

It might help if you look up specific heat capacity or similar on wikipedia..

http://en.wikipedia.org/wiki/Heat_capacity

Heat capacity (usually denoted by a capital C, often with subscripts), or thermal capacity, is the measurable physical quantity that characterizes the amount of heat required to change a substance's temperature by a given amount. In the International System of Units (SI), heat capacity is expressed in units of joule(s) (J) per kelvin (K).

Derived quantities that specify heat capacity as an intensive property, i.e., independent of the size of a sample, are the molar heat capacity, which is the heat capacity per mole of a pure substance, and the specific heat capacity, often simply called specific heat, which is the heat capacity per unit mass of a material.
 
  • #3
moimoi24
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Thank you for the reply...
I just got confused, when i solved it, i got a higher temperature when in fact it must be lower than the initial temperature. How is that? thanks for the help.
 
  • #4
moimoi24
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Is the the change in temperature denoted by final temperature - initial temperature? or vice versa? please help me...
 
  • #5
CWatters
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The equation gives you the change in temperature. Obviously if you put hot metal into cold water the temperature of the metal will fall...

Temperature change = Energy/ (Mass * specific_heat_capacity)

= 30000/(0.23 * 140)
= 932 K

If it starts at 1200 it will cool to 1200-932=268K

However water isn't liquid at 268K because that is -5C so there is a problem.

Assuming I haven't made a mistake..

Either the water was frozen when the metal was added or the problem has an error in it. The metal can't give up that much energy to water if the block started at 1200K and it was placed in liquid water. There wasn't that much energy in the metal block (relative to 0C).
 
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  • #6
moimoi24
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Thank you so much,,, what about using this formula, what made this one wrong? because if i use the i'll get 2132K as the final temperature.
Q = m*c*(Tf - Ti)

but if i use Q = m*c*(Ti-Tf), i'll get 268K? Can you help me explain this one...
Thanks.
 
  • #7
Neandethal00
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Thank you so much,,, what about using this formula, what made this one wrong? because if i use the i'll get 2132K as the final temperature.
Q = m*c*(Tf - Ti)

but if i use Q = m*c*(Ti-Tf), i'll get 268K? Can you help me explain this one...
Thanks.

I didn't read your whole problem, but I saw your last equation and dilemma.

If object is heating, Q is positive in Q=mc(Tf-Ti)
If object is cooling, losing heat, then Q is negative. -Q=mc(Tf-Ti).
 
  • #8
moimoi24
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I didn't read your whole problem, but I saw your last equation and dilemma.

If object is heating, Q is positive in Q=mc(Tf-Ti)
If object is cooling, losing heat, then Q is negative. -Q=mc(Tf-Ti).

thank you so much,, it really helps me because i really got confused... thank you so much...
 
  • #9
Vanadium 50
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Glad we could be of help, but in the future, please post questions like this in the Homework Help sections.
 

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