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Mathhelp346
I know the mass (.082 kg) and the ΔT (-65°C) of a metal. Using this, how can I find the heat capacity (J/kgC°) and heat (J)? Is it even possible?
My friend said that the heat is just the opposite of the water's heat, which was 2491 J. So, would the heat be -2491 J?
An object's heat capacity (symbol C) is defined as the ratio of the amount of heat energy transferred to an object to the resulting increase in temperature of the object. \[C =Q \frac{ Q }{ DeltaT }\] \[[C] = \frac{ J }{ Kg \times K° }\] For your case you might want to combine the formulas. Remember: Always convert to Kelvin when doing Physics, a step so easy to forget but so important for getting it right in your formulas. After that they might state that the answer should be in Celsius or other. Hope this helps.
What do you mean combine?
How would I combine them?
I trying to figure that out, it is getting late over here so I might have to get back to this tomorrow but I will try to keep it in mind and try to find a solution. Important : it is a typo in my top formula it should be: \[C =\frac{ Q }{ \Delta T }\]
Last attempt to throw you in the right direction before call the night here, hope it does not just confuse you. \[E _{k} = \frac{ 3 }{ 2 } k \times T\] Where \[k = 1,38 \times 10^{-23} J/K\] (better known as Boltzmann constant) This is to find the energy (between the molecules) the temperature is always a result of the energy.
Sorry that just confused me..
Formula \[c \times m \times \Delta T = Q\] to get to the heat capacity we just do some algebra to it ant get \[c = \frac{ Q }{ m \times \Delta T }\] Where C : heat capacity in J per Kg Q : heat in Joule ∆T:(is given in Kelvin degrees) The Boltzmann stuff above is to convert between joules and degrees celsius. Given the information in the questioin it is not possible due to that Q is unknown in the equation and with the information given it is no way of finding it. So if it in your assignment try finding out if there are some assumed values or some more info there, anyways it might be a trick question.