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)

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