# Heat capacity of pasta

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.42.

Here’s another example of heat capacity. In what must be blatant product placement, Schroeder tells us that the specific heat capacity of Albertson’s Rotini Tricolore pasta is about ${1.8\mbox{ J g}^{-1}\;^{\circ}\mbox{C}^{-1}}$. [Actually, I’m not sure this really exists, as I couldn’t find any mention of Albertson’s pasta. Rotini pasta is the same as fusilli, which is the helical pasta.] We put 340 g of this pasta at ${25^{\circ}\mbox{ C}}$ into 1.5 litres of boiling water. At what temperature ${T}$ does the system come to equilibrium?

The water transfers an amount of heat ${Q}$ to the pasta, so the water cools a bit and the pasta heats up. The specific heat capacity of water is ${1\mbox{ cal g}^{-1}\;^{\circ}\mbox{C}^{-1}=4.186\mbox{ J g}^{-1}\;^{\circ}\mbox{C}^{-1}}$. The total heat capacities are

 $\displaystyle C_{pasta}$ $\displaystyle =$ $\displaystyle 1.8\times340=612\mbox{ J }^{\circ}\mbox{C}^{-1}\ \ \ \ \ (1)$ $\displaystyle C_{water}$ $\displaystyle =$ $\displaystyle 4.186\times1500=6279\mbox{ J }^{\circ}\mbox{C}^{-1} \ \ \ \ \ (2)$

Therefore

 $\displaystyle \Delta T_{pasta}$ $\displaystyle =$ $\displaystyle T-25=\frac{Q}{C_{pasta}}=\frac{Q}{612}\ \ \ \ \ (3)$ $\displaystyle \Delta T_{water}$ $\displaystyle =$ $\displaystyle T-100=\frac{-Q}{C_{water}}=-\frac{Q}{6279} \ \ \ \ \ (4)$

Dividing the first equation by the second, we get

 $\displaystyle \frac{T-25}{100-T}$ $\displaystyle =$ $\displaystyle \frac{6279}{612}\ \ \ \ \ (5)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 93.3^{\circ}\mbox{C} \ \ \ \ \ (6)$

The heat transferred is

$\displaystyle Q=612\left(93.3-25\right)=4.18\times10^{4}\mbox{ J} \ \ \ \ \ (7)$