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How can we find specific heat capacity without using heat, $$ c= \dfrac{\delta Q}{m \,\delta T}$$ Calorimeter $$ c_1\cdot m_1\cdot \delta T_1 =c_2\cdot m_2 \cdot \delta T_2 $$ In this formula we can measure mass by balance and temperature with thermometer. But we need one specific heat either $c_1$ or $c_2$ to calculate other material heat capacity.

The question arise how do we know heat capacity of the first material to use as reference, because we can not calculate heat & specific heat capacity directly. How the first time we calculate heat capacity.

Is there any reference material heat capacity to use as relative. Or any statistical calculation or any other?

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$cm\delta T=\delta Q$ , that's pretty much self evident , for the reference material the $\Delta Q$ is calculated in isolation , maybe with a burner or something , by heating it , Now you already have the thermometer and from the amount of fuel exhausted you know the heat change for say in the temperature change $\Delta T$ . Now if we have a same temperature change in your two body setting then you already know the heat change in one of the body from the above experiment .So in the second equation you've put $c_1m_1\delta T$ is known because its equal to $\delta Q$ , you see there is no variable except $c_2$ now.
How is the calorific value of fuel calculated ?
The fuel is put in a bomb calorimeter , upon burning the fuel releases heat and heats up the surrounding water , of which we calculate the temperature change.The unit of heat is defined with respect to water (calorie units) as you must be knowing.Since you know the temperature change , of water , you directly know the number of calories of heat released by fuel(calorific value) , problem solved , now you can proceed with the first paragraph.

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  • $\begingroup$ sure , I will edit my answer to include that. $\endgroup$ – ADITYA PRAKASH Mar 24 at 7:25
  • $\begingroup$ Thanks for the reply @ADITYA PRAKASH . Can pls explain how do we measure amuont of heat and formula. because i think it can not be measure directly as temperature , pressure , volume , mass etc.. In an isothermal expansion or compression change in heat $$ \delta Q = \delta W = - n.R.T ln(V_2/V_1) = -n.R.T ln(P_1/P_2) $$ Does this process explain change in heat $ \delta Q $ . and ischoric / isobaric for $C_v$ and $C_p$ $\endgroup$ – 123 Mar 24 at 7:26
  • $\begingroup$ That's an approach too,but there's an easier approach with bomb calorimeter , a purely experimental approach. $\endgroup$ – ADITYA PRAKASH Mar 24 at 7:39
  • $\begingroup$ i have installed service and run bomb calorimeter, but in this we use reference material or one has to know the reference material heat capacity. In this formula $$ \ c(T) = (\delta Q) / m . \delta T $$ mass can be measured by balance and Temperature by thermometer and what about two unknown physical quantities $ C_v $ and $ \delta Q $ . if do not know the $ C_v $ how can we know $ \delta Q $ . If this is the formula how the very first time we know $ C_v $ to put the value in this formula $\endgroup$ – 123 Mar 24 at 7:49
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    $\begingroup$ Now i found the answer . my was hidden in "MECHANICAL EQUIVALENT OF HEAT" because we know mechanical energy and we did not know heat energy . we can find conversion of mechanical energy energy in heat. then we can find specific heat capacity. $\endgroup$ – 123 Mar 24 at 8:56
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I can't see your problem. Clearly you have to choose some substance as a standard for specific heat capacity. The same happens every time you enter a new field. E.g. it happened with electric charge when the coulomb wasn't yet there.

The first standard was liquid water. The calorie was defined stating that specific heat capacity of water was 1 cal/g/°C. (To be precise, temperature and pressure should be specified.)

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