PRODUCT LOAD Mass of ice at $-4{ }^{\circ} \mathrm{C}$ is needed to cool $115 \mathrm{~kg}$ of vegetables in a bunker for 24 hours. The initial temperature of vegetables is assumed to be $30^{\circ} \mathrm{C}$. It is also assumed that the average temperature inside the bunker is $7^{\circ} \mathrm{C}$ within 24 hour period. If the heat gained per hour in the bunker is $30 \%$ of the heat removed to cool the vegetable from $30^{\circ} \mathrm{C}$ to $7^{\circ} \mathrm{C}$, what would be the required mass of ice? Note: Specific Heat of Vegetable $=3.35 \mathrm{~kJ} / \mathrm{kg}-\mathrm{K}$ Specific heat of ice $=1.935 \mathrm{~kJ} / \mathrm{kg}-\mathrm{K}$ Specific Heat of Water $=4.186 \mathrm{~kJ} / \mathrm{kg}-\mathrm{K}$ Heat of fusion of ice $=335 \mathrm{~kJ} / \mathrm{kg}$

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