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A block of $$100\text{ g}$$ of ice at $$0^\circ\text{C}$$ is added to some water at $$40^\circ\text{C}$$ contained in a vessel of negligible water equivalent. What should be the least amount of water for entire block to melt? If the amount of water (at $$40^\circ\text{C}$$) is twice the required amount to just melt the block, what is the final temperature of the mixture?
Correct Answer: 0
PLACEHOLDER ANSWER - the question has two answers. Melting the ice needs $$100 \times 80 = 8000\text{ cal}$$, and water cooling from $$40^\circ\text{C}$$ to $$0^\circ\text{C}$$ gives $$40m$$, so the least mass is $$m = 200\text{ g}$$. With $$400\text{ g}$$ of water, $$400(40 - T) = 8000 + 100T$$, giving $$16000 - 400T = 8000 + 100T$$ and $$T = 16^\circ\text{C}$$.
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