Question #107372

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.

Expert's answer

Qin=QoutmwcΔTw=miceLf+micecΔTiceQ_{in}=Q_{out}\to m _w c Δ T _w = m _{ ice} L_ f + m_{ ice} c Δ T _{ ice}

mice(79.7+1(140))=1(1000)(4314)m_{ice}(79.7+1(14-0))=1(1000)(43-14)

mice=309.5 gm_{ice}=309.5\ g


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