You make some iced tea by dropping 154 g of ice into 500.0 mL of warm tea in an

Question

You make some iced tea by dropping 154 g of ice into 500.0 mL of warm tea in an insulated pitcher If the tea is initially at 18.4 degree C and the ice cubes are initially at 0.00 degree C, how many grains of ice will still be present when the contents of the pitcher reach a final temperature? The tea is mostly water, so assume that it has the same density (1 00 g mL), molar mass, heat capacity (75.3 J K^-1 mol^-1), and heat of fusion (6.00 kJ mol) as pure water. The heat capacity of ice is 37.7 J K^-1 mol^-1.

Explanation / Answer

We have 500g or (500/18)= 27.8 mol water (in the form of tea) at 18.4oC which needs to be converted into ice water at 0oC. We need to cool water by 18.4oC or 18.4K. This cooling energy will be given by fusion of ice. So, energy of cooling of water should be equal to heat of fusion of ice as the container is insulated and no energy is being exchanged with the surroundings.

Energy evolved by cooling of water = Heat capacity X change in temperature X number of moles of water

= 75.3 JK-1mol-1 X 18.4 K X 27.8 mol

= 38517 J or 38.517 kJ

This will cause fusion or melting of a certain number of moles of ice. Let that number be x.

Total heat of fusion of ice = Heat of fusion of ice X number of moles

= 6 kJ/mol X (x) mol = 6x

By logic,

6x = 38.517

x= 6.42 mol

That is 6.42 mol or (6.42 X 18)= 115.56g of ice will be converted to water at 0oC

The remaining amount = 154-115.56 = 38.4g ice will remain in the insulated pitcher at the final temperature.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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