91Ó°ÊÓ

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

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

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

Now find the change in temperature using

$C=\frac{Q}{\mathrm{Δ}T}$

For water

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

Similarly for Pasta

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

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