Calorimetry Problem I was doing a problem in thermodynamics where the net heat is 0.
I don't understand why if you have say a copper calorimeter with water at say 15 °C and add a mass of copper at a higher temperature say 90 °C that when calculating the final temperature you would use for the copper piece this in the formula:
$$ Q = mc(T_{f}-T_{i}) $$
Where f is for final and i is for initial. Say mass is 0.3 kg. I was told that regarding $ T_{f}-T_{i} $ I would need to use $ 90 - T $.
This problem was to work out the final temperature of the system. Why was the initial 90 °C of the added copper substituted with $ T_{f} $ in the heat equation?
 A: When you use 90-T, you are determining the amount of heat that the copper lost.  When you use T-90, you are determining the amount of heat that the copper gained, which is the negative of how much it lost.  So, it depends on how you are setting up the equation.  If you put the heat gained by the water equal to the heat lost by the copper, that is one version of the set up.  If you put the sum of the heat gained by the water plus the heat gained by the copper on the same side of the equation, and are setting that equal to zero, that's an equivalent version of the set up.  The second version is the same as your "net heat is 0."
A: Heat lost by copper calorimeter = Heat gained by water
$-mc(T_f-T_i)=m_wc_w(T_f-T_i)$
For thermal equilibrium final temperature for both the components will be the same. Let the equilibrium temperature be T and $15^{\circ}C\leq T\leq90^{\circ}C$
$-mc(T-90)=m_wc_w(T-15)$
$mc(90-T)=m_wc_w(T-15)$
Its quite easy. When a hot object is kept in contact with a cooler object, the hot thing will lose heat and the cold thing will gain that energy. Therefore, for the hot object $\Delta{Q}=-ve$ and for the cold object $\Delta Q=+ve$.
To avoid confusion just use the magnitudes of their respective change in heat.
A: This is the concept based on $\text{heat gained} = \text{heat lost}$.we basically have $$m_1 c_1 (\delta T_1) = m_2 c_2 (\delta T_2)$$ we are now concerned of either heat lost or heat gained. By $(90-T)$ we mean that the temperature of copper has gone down. And by $(T-90)$ we mean the temperature of the body has gone up. That's the difference. Hope it was helpful
A: In going along with Mitchell's idea of avoiding confusion, I always set this problem up exactly as stated.  Since heat is equivalent to energy, it is my opinion that it should always have a positive value.  Accordingly, the heat lost by the copper is equal to $m_c C_c (T_i - T_f)$, while the heat gained by the water is equal to $m_w C_w (T_f - T_i)$.  When these two quantities are set equal to each other, a bit of algebra yields an equation that solves the general calorimetry problem for an unknown final temperature.
