You have hot water of a mass Minitially at 80°C. You add ice, initially at 0°C,...

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You have hot water of a mass Minitially at 80°C. You add ice, initially at 0°C, to bring the mixture to 20°C. Which of the fo

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Answer 1

Answer #1

Solution:

At equilibrium,

Heat added = heat removed

$m_{ice}\left ( C_{ice}\times \left (20^{o}C-0^{o} C \right )+L_{ice} \right )=MC_{water}\times \left ( 80^{o}C-20^{o}C \right )$

$m_{ice}=M\frac{\left (C_{water}\times \left ( 80^{o}C-20^{o}C \right ) \right )}{\left ( C_{ice}\times \left (20^{o}C-0^{o} C \right )+L_{ice} \right )}$

$m_{ice}=M\left (\frac{C_{water}\times 60^{o}C }{ C_{ice}\times 20^{o}C+L_{ice} } \right )$

Answer : option 3

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Answer 2

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