One liter of water at 43◦C is used to make iced
tea.
How much ice at 0 ◦C must be added to
lower the temperature of the tea to 14 ◦C?
The specific heat of water is 1 cal/g · ◦ C and
latent heat of ice is 79.7 cal/g.
Answer in units of g.
Qin=Qout→mwcΔTw=miceLf+micecΔTice
mice(79.7+1(14−0))=1(1000)(43−14)
mice=309.5 g
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