I need to find the temperature two things equilibrate to in a constant pressure calorimeter given only their specific heats, masses, and starting temps.

The only thing I know for specific heat is that. $c = \frac{J}{G\times \Delta T}$, and how to work that around to find if one is missing, and well as $\frac{q}{\Delta T}$ for doing things with calorimeters, though a different kind. That's the only equation I know for this stuff. In this question I have neither the energy nor the change in temp. I know how to solver it if the two substances have equal specific heats, then you just find the mid point between temperatures. But two different substances, I have no idea. I tried to do it the hard way of just slowly taking of/adding one *C of each and matching up the energies required but that got me nowhere.

  • $\begingroup$ I really wish people would comment on down voting and explain... If I knew what the problem was already I wouldn't have done it.... $\endgroup$ – Cr157 Oct 5 '14 at 6:05
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    $\begingroup$ Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. That's what someone was likely reacting to with their downvote. I know you may be frustrated with the question, but if you put something forth (what equations do you know for specific heat, for example?) someone can steer you in the right direction. $\endgroup$ – jonsca Oct 5 '14 at 6:07
  • $\begingroup$ Do you have the specific heat of the calorimeter given to you? If not, is it made of something that doesn't absorb a lot of heat? $\endgroup$ – jonsca Oct 5 '14 at 6:45
  • $\begingroup$ No. I says constant pressure calorimeter, and according to the book I don't have to worry about the amount of heat that type absorbs because it's really small. $\endgroup$ – Cr157 Oct 5 '14 at 6:54

Assuming the heat absorption of the calorimeter is negligible (so if it's made of an insulating material, the only changes in heat will be of the contents themselves).

Substance 1 (the one at the higher initial temperature) has a heat = $q_1$

Substance 2 has a heat of $q_2$

Since substance 1 will lose heat until it comes to equilibrium with substance 2, the heat lost by 1 will be absorbed by 2, so $q_1 = -q_2$

If $q_1 = m_1c_{p1}(T_{{1}_{initial}} - T_{{1}_{final}})$ and $q_2 = m_2c_{p2}(T_{{2}_{initial}} - T_{{2}_{final}})$ and $T_{1_{final}} = T_{2_{final}} = T_f$ you can solve for $T_f$.

If the heat absorption of the calorimeter was not negligible, then $q_1 + q_2 + q_{cal} = 0$

  • $\begingroup$ I get it if I had q1 or q2, but I don't. How in the word do you that with just m1xcp1x? m2xcp2x? $\endgroup$ – Cr157 Oct 5 '14 at 7:10
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    $\begingroup$ @Cr157 You have the masses, the specific heats at constant pressure ($c_p$), and the initial temperature of each, so you can fill in $m_1, m_2$ and the specific heats, make the two equal and solve for $T_f$. $\endgroup$ – jonsca Oct 5 '14 at 7:14
  • $\begingroup$ @Cr157 See if that makes it a bit more clear. $\endgroup$ – jonsca Oct 5 '14 at 7:17
  • $\begingroup$ Not there yet... but I'm trying to get it... $\endgroup$ – Cr157 Oct 5 '14 at 7:19
  • $\begingroup$ @Cr157 $q_1 = -q_2$, so it's just a matter of algebra at this point with $T_f$ as the unknown. $\endgroup$ – jonsca Oct 5 '14 at 7:25

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