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)?

Short answer

Final temperature is 366.48 K (93.48^oC )

Step-by-step solution

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

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

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,

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

Temp will be increased to 93.48^oC