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 248 grams of lemonade at 25.7°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 -10.5°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 = 2.06 J/g·°C Heat of fusion = 333 J/g

Answer 1

To cool the lemonade, Chris will need to use 448.7 grams of ice.

Step-by-step explanation:

To determine the amount of ice used to cool the lemonade, we can use the principle of energy conservation. The heat lost by the lemonade is gained by the ice, causing it to melt and reach the same temperature. The equation for energy conservation is:

heat gained by ice = heat lost by lemonade

We can calculate the heat gained by the ice by using the equation:

q = m * c * ΔT

Where q is the heat gained or lost, m is the mass, c is the specific heat, and ΔT is the change in temperature. Given that the specific heat of ice is 2.06 J/g·°C and the ice starts at -10.5°C and reaches 11.2°C, we can calculate the mass of the ice as:

m = q / (c * ΔT)

Plugging in the values, we find that Chris would need to use 448.7 grams of ice.