# How do I find the final temperature of a calorimeter using specific heat capacities?

What is the final temperature of water and iron if a $$\pu{30 g}$$ piece of iron at $$\pu{144 °C}$$ was dropped into a calorimeter with $$\pu{40 g}$$ of water at $$\pu{20 °C}$$? Specific heat of water is $$\pu{4.184 J g-1 ^\circ C-1}$$, and of iron is $$\pu{0.449 J g-1 ^\circ C-1}$$

Here is my work: \begin{align} Q &= mc\,\Delta T\\ Q_1 &= (\pu{30 g})(\pu{0.449 J g-1 ^\circ C-1})(x - \pu{144 ^\circ C}) \tag{Iron}\\ Q_2 &= (\pu{40 g})(\pu{4.184 J g-1 ^\circ C-1})(x - \pu{20 ^\circ C}) \tag{Water}\\ \text{Since,} \quad Q_1 &= -Q_2\\ 13.47 (x-144) &= - (167.36)(x-20)\ \pu{J}\\ 13.47x - 1939.68 &= -167.36x + 3347.20\\ 180.83x &= \pu{5286.88 ^\circ C} \\ x &= \pu{0.03420 ^\circ C} \end{align}

This is giving me an answer that isn't correct according to my book. What did I do wrong, and how can I fix it?

• $\frac {5286.88}{180.83} \neq 0.03420$ – LDC3 Apr 20 '14 at 2:48
• Use Kelvin instead of Centigrade/ Celsius! It would not change in this calculation as they are on the same scale and you are using differences. Also try using units throughout the whole process, this will give you a hint if you transformed your equations correctly. Apart LDC3's comment, I cannot see anything wrong. – Martin - マーチン Apr 20 '14 at 9:23

All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius. \begin{align} Q &= mc\Delta T\\ \end{align} Now you can form the equations for each of the problem, while substituting $\Delta T$ with a temperature range, being $x$ the final temperature the whole system will end up on. Also note, that the iron will be cooled down, while the water will be heated. (I am using a different approach than you. \begin{align} Q_\mathrm{loss} &= m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x]\\ Q_\mathrm{gain} &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\ \end{align}
The transferred heat has to equal $$Q_\mathrm{gain} = Q_\mathrm{loss}$$
With this you can solve for $x$. \begin{align} m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x] &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\ %m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe})-m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x-m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})\\ % m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x+m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x\\ % m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= [m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})]\cdot{}x\\ x &=\frac{m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})}{ m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})}\\ % x &=\frac{30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}\cdot{}417~\mathrm{K} +40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}\cdot{}293~\mathrm{K}}{ 40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}+30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}}\\ x &= \frac{5616.99~\mathrm{J}+49036.48~\mathrm{J}}{167.36~\mathrm{\frac{J}{K}}+13.47~\mathrm{\frac{J}{K}}}\\ x &= \frac{54653.47}{180.83}~\mathrm{K} =302.24~\mathrm{K}\\ x &\approx 29~\mathrm{^\circ{}C} \end{align}