What would be the temperature of 40.0 g of 22°C water mixed with 60.0 g of 80°C water? Answer in degrees Celsius. Here's the logical pathway to find the answer: The cold water absorbs heat according to Q1 = 40*4.186*(T-22). The hot water loses heat according to Q2 = 60*4.186*(80-T). Q1 must equal Q2, so solve for T, the equilibrium temperature.

What would be the temperature of 40.0 g of 22°C water mixed with 60.0 g of 80°C water? Answer in degrees Celsius. Here's the logical pathway to find the answer: The cold water absorbs heat according to Q1 = 404.186(T-22). The hot water loses heat according to Q2 = 604.186(80-T). Q1 must equal Q2, so solve for T, the equilibrium temperature.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter19: Temperature, Thermal Expansion And Gas Laws

Section: Chapter Questions

Problem 22PQ

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