The weight of heat

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  • #1
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I read that if the temperature of 1 Kg of gold rises by 10°C its mass increases by 1.4 * 10^-14 kg.
Can you please explain how you get such value from E= mc^2?
I think the increase corresponds to E = h f =1.53*10*10^36 and since
1 kg has 1.1*10^30 electrons, each one gets 1.4*10*6 h
but
10 degrees correspond to energy E = h f = 2.8*10^11 h)

so shouldn't the increase of energy be 2.29*10 ^41 h?

Can you also tell what instrument can measure the weight o 10^-14 Kg and what is the minimal weight that can be detected?
 

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  • #2
Orodruin
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It is completely unclear what you are trying to do. E=hf is the energy content of a photon and has absolutely nothing to do with the mass increase of the gold. You need to figure out how much energy needs to be added and compute the corresponding mass increase.
 
  • #3
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It is completely unclear what you are trying to do. E=hf is the energy content of a photon and has absolutely nothing to do with the mass increase of the gold. You need to figure out how much energy needs to be added and compute the corresponding mass increase.
E = 1.4*10-14* 9*10^16 = 12 = 1260 J
The energy of a joule can be expressed in many ways, in eV or in h*Herz 1260 /h = 1.9*10 ^36
what is the problem? that is the energy added, I cannot figure out how that relates to T = +10*C

Any ideas?
 
  • #4
Orodruin
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The energy of a joule can be expressed in many ways, in eV or in h*Herz 1260 /h = 1.9*10 ^36
what is the problem?
Which units you use are unrelated to how you solve the problem. There is no point in using E=hf here.

E = 1.4*10-14* 9*10^16 = 12 = 1260 J
Just putting numbers is not a very good way to present a solution. First write down the physical relation you are using, then write down what numbers you are putting in (including units!).

that is the energy added, I cannot figure out how that relates to T = +10*C
The energy added must depend on the temperature increase. If yours does not, it is wrong.
 
  • #6
Doc Al
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To understand how to calculate the "heat" energy required to raise the temperature of something, read this: Specific Heat
 
  • #7
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Thank you very much for the links, so:
c for gold is 0.126 J/gmK 1000g, therefore 1000g*10 K makes 1260 J (1.5*10^36 hv), right?
 
  • #8
Doc Al
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c for gold is 0.126 J/gmK 1000g, therefore 1000g*10 K makes 1260 J
1260 J looks good to me.

(1.5*10^36 hv)
What's this?
 
  • #9
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1260 J looks good to me.
What's this?
Is it so weird to refer to energy by Hz? 1 J equals 1.5*10^33 h/s, right?

Can you tell me if the weight increase is just deduced from the formula, or if it is (and to what precison) confirmed my measurement?
 
  • #10
Orodruin
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Is it so weird to refer to energy by Hz?

It simply does not make any sense. You could do this by using units of ##\hbar## but this is normally done in particle physics and this is not.
Also, in your answer you are referring to ##hf##. This makes no sense at all unless you specify what ##f## is.
 
  • #11
jtbell
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I read that if the temperature of 1 Kg of gold rises by 10°C its mass increases by 1.4 * 10^-14 kg.
Can you please explain how you get such value from E= mc^2?
c for gold is 0.126 J/gmK 1000g, therefore 1000g*10 K makes 1260 J
So now you know E (or more precisely, the amount by which the E of the gold increases when its temperature increases by 10°C). Now, what m does this give you?
 
  • #12
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This makes no sense at all unless you specify what ##f## is.
You must have missed that, in my first post I specified
f =1.53*10*10^36
 
  • #13
Orodruin
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You must have missed that, in my first post I specified
This is completely besides the point. You cannot just arbitrarily select a frequency.

Apart from that, your "frequency" is not a frequency, it is a number. A number is meaningless without the appropriate physical units.
 

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