# Transfer of energy

After completing a lab for my chemistry class I am finding it difficult to find the initial temperature of an iron nail after being heated.

Procedure of the lab went as follows: heat an iron nail, place the heated nail in room temperature water, and record the temperature of the water.

The data collected from the lab: Given 1. Specific heat of water-1cal/gc 2. Specific heat of iron-.113cal/gc

Water 1. Volume-75ml 2. Mass-75g 3. initial temp-23c 4. final temp-27c

Nail 1. Mass-2.51g 2. Initial temp-x 3. Final temp-27c

So as I stated previously I have not been successful in finding the initial temp of the Iron nail based off of the given data.

I had tried using (mcΔt)w=(mcΔt)n being (75)(1)(27-23)=(2.51)(.113)(27-X) in the logic of the temperature of the nail and water reaching equilibrium but by using that formula x comes out to be -1029.338c which is most obviously not the initial temperature of the nail.

• As far as I can tell, all of the math is correct. However, 75mL of water is a lot for such a small nail. Are you sure you have the mass of the nail and the water correct? – Shafter Nov 13 '14 at 7:49
• It was the amount we were told to use by our instructor – Bailey Tincher Nov 13 '14 at 11:23

\begin{align}(mc\Delta T)_{\mathrm{water}} &= (75\mathrm{\ g})\left(1\ \frac{ \mathrm{cal}}{\mathrm{g\cdot °C}}\right)(27 - 23 \mathrm{°C}) \\ q &= 300\mathrm{\ cal} \end{align}
\begin{align} \Delta q_\mathrm{sys} &= q_\mathrm{water}+q_\mathrm{nail}\\ 0 &= q_\mathrm{water}+q_\mathrm{nail}\\ q_\mathrm{water} &= -q_\mathrm{nail}\end{align}
$\Delta q_{sys}$ is zero because we have a (approximately) closed system where no heat can enter or leave. If you substitute $mc\Delta T$ for q, you'll see that you get an equation very similar to the one you used, only with an added negative sign. Because water is gaining heat, the nail must be losing it, so the signs are opposite.