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 201grams of lemonade at 23.8°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 10.5°C. Of course, the ice will melt and reach the same temperature. If the ice starts at -14.3°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.

__g

Data for water at 1 atm: Melting point = 0.0°C Specific heat liquid = 4.18 J/g·°C Specific heat solid = 2.06 J/g·°C Heat of fusion = 333 J/g

Explanation / Answer

-Qhot = Qcold

-mwater*Cpwater*(Tf-Thot) = m*ice*(Tf-Ti) + m*Latentheat + m*Cpwater*(Tf-Ti)

-201*4.184*(10.5-23.8) = m*2.06*(0--14.3) + m*333+ m*4.184*(10.5-0)

solve for m

11185.08 = m*(2.06*14.3 + 333 + 4.184*10.5)

m = 11185.08 /406.39

m = 27.523 g of ice required

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.