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$1 \mathrm{Kg}$ of ice at $-20^{\circ} \mathrm{C}$ is mixed with $2 \mathrm{Kg}$ of water at $90^{\circ} \mathrm{C}$. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume latent heat of ice $=334.4 \mathrm{KJ} / \mathrm{Kg}$, specific heat of water and ice are $4.18 \mathrm{~kJ} /(\mathrm{kg} \cdot \mathrm{K})$ and $2.09 \mathrm{~kJ} /(\mathrm{kg} \cdot \mathrm{K})$, respectively.)
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The correct answer is:
$30^{\circ} \mathrm{C}$
$m_{i} s_{i}(\Delta T)+m_{i} L+m_{i} s_{w}(T-0)=m_{w} s_{w}(90-T)$
$1 \times 2.09(20)+1 \times 334.4+1 \times 4.18 \times T=2 \times 4.18 \times(90-T)$
$T=60-30=30^{\circ} \mathrm{C}$
$1 \times 2.09(20)+1 \times 334.4+1 \times 4.18 \times T=2 \times 4.18 \times(90-T)$
$T=60-30=30^{\circ} \mathrm{C}$
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