A person makes a quantity of iced tea by mixing 740 g of hot tea (essentially wa

Question

A person makes a quantity of iced tea by mixing 740 g of hot tea (essentially water) with an equal mass of ice at its melting point. Assume the mixture has negligible energy exchanges with its environment. (a) If the tea's initial temperature is Ti = 81°C, when thermal equilibrium is reached what are the mixture's temperature Tf and (b) the remaining mass mf of ice? If Ti = 62°C, when thermal equilibrium is reached what are (c) Tf and (d) mf? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg.

Explanation / Answer

)
m = 740 g = 0.74 kg

C= 4186 J/(kg K)

Lf = 333*10^3 J/kg

Apply, Heat lost by water = heat ganined by ice

m*c*(Ti - Tf) = m*Lf + m*c*(Tf - 0)

c*(Ti - Tf) = Lf + c*Tf

c*Ti - c*Tf = Lf + c*Tf

C*Ti - Lf = 2*c*Tf

Tf = (c*Ti - Lf)/(2*c)

= (4186*81 - 333*10^3)/(2*4186)

= 0.724 C

b) total ice melts
so, mf = 0

c) Tf = 0

d) mass of ice that melt, m = m*c*(62-0)/Lf

= 0.74*4186*62/(333*10^3)

= 0.577 kg

so, ramaining mass of ice, mf = 0.74 - 0.577

= 0.163 kg

= 163 grams

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