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$5 \mathrm{~g}$ of ice at $-30^{\circ} \mathrm{C}$ and $20 \mathrm{~g}$ of water at $35^{\circ} \mathrm{C}$ are mixed together in a calorimeter.
The final temperature of the mixture is (Neglect heat capacity of the calorimeter, specific heat capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hea: capacity of water $\left.=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)$
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The final temperature of the mixture is (Neglect heat capacity of the calorimeter, specific heat capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hea: capacity of water $\left.=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)$
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Verified Answer
The correct answer is:
$9^{\circ} \mathrm{C}$
Given, $s_w=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
$\begin{aligned} & s_{\text {ice }}=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1} \\ & L_f=80 \mathrm{cal} \mathrm{g}^{-1}\end{aligned}$
Mass of ice, $m_{\text {ice }}=5 \mathrm{~g}$
Mass of water, $m_w=20 \mathrm{~g}$
Let final temperature of the mixture be $T^{\circ} \mathrm{C}$.
According to principle of calorimetry,
Heat lost by water $=$ Heat gained by ice
$\begin{aligned} m_w s_w(35-T)=m_{\mathrm{ice}} s_{\mathrm{ice}}[0- & (-30)] \\ & +\mathrm{m}_{\text {ice }} \times L_f+m_{\mathrm{ice}} \times s_w \times T\end{aligned}$
$\Rightarrow 20 \times 1 \times(35-T)=5 \times 0.5 \times 30+5 \times 80+5 \times 0.5 \times T$
On solving, we get
$T=9^{\circ} \mathrm{C}$
$\begin{aligned} & s_{\text {ice }}=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1} \\ & L_f=80 \mathrm{cal} \mathrm{g}^{-1}\end{aligned}$
Mass of ice, $m_{\text {ice }}=5 \mathrm{~g}$
Mass of water, $m_w=20 \mathrm{~g}$
Let final temperature of the mixture be $T^{\circ} \mathrm{C}$.
According to principle of calorimetry,
Heat lost by water $=$ Heat gained by ice
$\begin{aligned} m_w s_w(35-T)=m_{\mathrm{ice}} s_{\mathrm{ice}}[0- & (-30)] \\ & +\mathrm{m}_{\text {ice }} \times L_f+m_{\mathrm{ice}} \times s_w \times T\end{aligned}$
$\Rightarrow 20 \times 1 \times(35-T)=5 \times 0.5 \times 30+5 \times 80+5 \times 0.5 \times T$
On solving, we get
$T=9^{\circ} \mathrm{C}$
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