Ice cubes from the freezer weigh about 25 g and are at a temperature of -15.0 °C. Suppose I have a glass of ice tea (250 mL, 250.0 g), which is essentially all H₂O, at room temperature (22.0 °C) and I want to use ice cubes to cool it down to the temperature of a refrigerator (4.0 °C, 40 °F). How many ice cubes do I need to use? In other words, what mass of ice cubes are needed to get the combined system to 4.0 °C?

Answered: Ice cubes from the freezer weigh about…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Ice cubes from the freezer weigh about 25 g and are at a temperature of -15.0 °C. Suppose I have a glass of ice tea (250 mL, 250.0 g), which is essentially all H₂O, at room temperature (22.0 °C) and I want to use ice cubes to cool it down to the temperature of a refrigerator (4.0 °C, 40 °F). How many ice cubes do I need to use? In other words, what mass of ice cubes are needed to get the combined system to 4.0 °C?

Expert Solution

Step 1

The water in the glass will be in equilibrium with the ice until heat transformation occurs from ice to water to reach a common temperature 4 $\circ C$ .

Both ice and water have a specific heat constant = 4 $J / g / \circ C$

$∆ H_{i c e} = ∆ H_{i c e t e a} m_{i c e} \times s \times (t_{2} - t_{1}) = m_{i c e t e a} \times s \times (t_{1} - t_{2}) (t_{1} = 4^{\circ} C) m_{i c e} \times 4.2 \times (4 + 15) + \frac{}{} = 250 \times 4.2 \times (22 - 4) m_{i c e} = 236.84 g .$

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