It is a hot summer day and Chris wants a glass of lemonade. There is none in the

Question

It is a hot summer day and Chris wants a glass of lemonade. There is none in the refrigerator, so a new batch is prepared from freshly squeezed lemons. When finished, there are 246 grams of lemonade at 26.3°C. That is not a very refreshing temperature, so it must be cooled with ice. But Chris doesn't like ice in lemonade! Therefore, just enough ice is used to cool the lemonade to 11.2°C. Of course, the ice will melt and reach the same temperature. If the ice starts at -13.4°C, and if the specific heat of lemonade is the same as that of water, how many grams of ice does Chris use? Assume there is no heat transfer to or from the surroundings. Data for water at 1 atm: Melting point-0.0°C Specific heat liquid- 4.18 J/g.C Specific heat solid- 2.06J/g. Heat of fusion 333 J/g

Explanation / Answer

Q = heat change for conversion of ice at -13.4 oC to ice at 0 oC + heat change for
conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 11.2 oC

Amount of heat absorbed , Q = mcdt + mL + mc'dt
= m(cdt + L + c'dt' )
Where
m = mass of ice= ?
c' = Specific heat of water = 4.18 J/g degree C
c = Specific heat of ice= 2.06 J/g degree C
L= Heat of fusion of ice = 333 J/g
dt' = 11.2 -0 = 11.2oC
dt = 0-(-13.4)=13.4 oC
Plug the values we get Q = m(cdt + L + c'dt ')=407.4*m

Which must be equal to heat lost by lemonade= m"c"dt"

407.4*m= 246g*4.18(J/goC)*(26.3-11.2)

m= 63.3 g

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