Q. 1.42 - An Introduction to Thermal Physics

Question

The specific heat capacity of Albertson's Rotini Tricolore is approximately 1.8 J/g ^oC . Suppose you toss 340 g of this pasta (at 25^oC ) into 1.5 liters of boiling water. What effect does this have on the temperature of the water (before there is time for the stove to provide more heat)?

Step-by-Step Solution

Answer

Final temperature is 366.48 K (93.48^oC )

Step 1: Given information

Specific heat capacity of water, c = 1 cal/gK = 4.186J/gK
Specific heat capacity of pasta, c_pasta=1.8 J/gK
Mass of water, m_Water=1500 g
Mass of pasta, m_Pasta= 340 g
Initial temperature of water =100^oC=373.15 K
Initial temperature of pasta =25^oC = 298.15K

Step2:Explanation

We know heat capacity is given as 

C = m c

where m= mass and c= specific heat capacity

Find the heat capacity of Pasta and Water

$$\begin{gathered} \mathrm{For} \mathrm{Water} \\ {C}_{\mathit{\text{water }}}\mathit{=}{m}_{\mathit{\text{water }}}\mathit{\times }{c}_{\mathit{\text{water }}} \\ =(1500 g)\times (4.186 J/gK) \\ =6279 \mathrm{J}{ \mathrm{K}}^{-1}................................\left(1\right) \\ \mathrm{For} \mathrm{Pasta}, \\ {C}_{\text{pasta }}=m_{\text{pasta }}\times c_{\text{pasta }} \\ =\left(340g\right)\times (1.8 J/gK) \\ =612{ \mathrm{J}}^{-1}{ \mathrm{K}}^{-1}.......................................\left(2\right) \end{gathered}$$

Now find the change in temperature using 

$C=\frac{Q}{\Delta T}$

For water

$$\begin{gathered} C_{\text{water }}=\frac{Q_{\text{water }}}{\Delta T_{\text{water }}} \\ 6279 \mathrm{J}·{\mathrm{g}}^{-1}{ \mathrm{K}}^{-1}=\frac{Q_{\text{water }}}{\Delta T_{\text{water }}}......................\left(3\right) \end{gathered}$$

Similarly for Pasta

$$\begin{gathered} C_{\text{pasta }}=\frac{Q_{\text{passa }}}{\Delta T_{\text{pasta }}} \\ 612 \mathrm{J} g^{-1}{ \mathrm{K}}^{-1}=\frac{Q_{\text{pasta }}}{\Delta T_{\text{pasta }}}...........................\text{ (4) } \end{gathered}$$

Assuming no heat is lost anywhere else. 

Heat lost by water is equal to heat gain by Pasta.

$Q=Q_{\text{pasta }}=-Q_{\text{water }}$

From the equation (3) and (4)

$$\begin{gathered} 6279 \mathrm{J}·{\mathrm{g}}^{-1}{ \mathrm{K}}^{-1}=\frac{-Q}{T-373.15}...........................\left(5\right) \\ 612 \mathrm{J}·{\mathrm{g}}^{-1}{ \mathrm{K}}^{-1}=\frac{Q}{T-298.15}.............................\left(6\right) \end{gathered}$$

Solve for T by dividing (5) by (6), we get,

$$\begin{gathered} \frac{6279 J.g^{-1}{K}^{-1}}{612 J.g^{-1}{K}^{-1}}=\frac{-Q/(T-373.15K)}{Q/(T-298.15K)} \\ \frac{T-298.15 K}{373.15 K-T}=10.26 \\ T-298.15=3828.51-10.26T \\ 11.26T=4126.66 \\ T=366.48 \mathrm{K} \end{gathered}$$

Temp will be increased to 93.48^oC