In an insulated container, a 5.0g sample of water at 80. C was added to an unkno

Question

In an insulated container, a 5.0g sample of water at 80. C was added to an unknown mass of cooler water at 30. C. The final temperature of the resulting mixture became 25 C. what mass of cooler water was present? (Assume the container itself absorbed a negligible amount of heat.)

I know the correct answer is A. 45g ,but would like to know the steps to solve the question. Thanks!

12. In an insulated container, a 5.0 g sample of water at 80. C was added to an unknown mass of cooler water at 30. C. The final temperature of the resulting mixture became 35 C. What mass of cooler water was present? (Assume the container itself adsorbed a negligible amount of heat.) 9-MCAT A 45 g 10. g 3s C. 2.5g D. 50. g E. 5.0g

Explanation / Answer

Given data

Assume Cp is constant

Mass of hot water mh = 5 g

Initial temperature of hot water T1 = 80 °C

Mass of cold water mc =?

Initial temperature of cold water T2 = 30 °C

Final temperature T = 35°C

Heat lost from the hot water = heat gained by the cold water

mh x Cp (T1 - T) = mc x Cp x (T - T2)

5 g x (80-35) = mc x (35 - 30)

mc = 45 g

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