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One kilogram of ice at $0^{\circ} \mathrm{C}$ is mixed with one kilogram of water at $80^{\circ} \mathrm{C}$. The final temperature of the mixture is
(Take : specific heat of water $=4200 \mathrm{~J}^{-1} \mathrm{~K}^{-1}$, latent heat of ice $=336 \mathrm{~kJ}^{-1} \mathrm{~kg}^{-1}$ )
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(Take : specific heat of water $=4200 \mathrm{~J}^{-1} \mathrm{~K}^{-1}$, latent heat of ice $=336 \mathrm{~kJ}^{-1} \mathrm{~kg}^{-1}$ )
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Verified Answer
The correct answer is:
$0^{\circ} \mathrm{C}$
$\begin{aligned} & \theta_{\text {mix }}=\frac{m_w \theta_w-\frac{m_i L_i}{c_w}}{m_i+m_w} \\ & \Rightarrow m_i=m_w \Rightarrow \theta_{\text {mix }}=\frac{\theta_w-\frac{L_i}{c_w}}{2}=\frac{80-\frac{336}{4.2}}{2}=0^{\circ} \mathrm{C}\end{aligned}$
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