A. Heat Transfer and Temperature Change 1. Using the calorimetry simulator, determine the final temperature of a mixture of water when 110 grams of water at 90°C was added to 130 grams of water at 20°C.

College Physics
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ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
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Chapter14: Heat And Heat Transfer Methods
Section: Chapter Questions
Problem 22PE: To help prevent from damage, 4.00 kg at 0C water is sprayed onto a fruit tree. (a) How much heat...
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A. Heat Transfer and Temperature Change
1. Using the calorimetry simulator, determine the final temperature of a mixture of water when 110 grams of water at 90°C was added to 130 grams of water at 20°C.
2. Perform manual calculations to find the final temperature using the equation: (specific heat of water = 4.184J/g.°C)
3. Compare the results with the reading on simulator.
4. Calculate the amount of heat transferred in the process.
heat given off by hot water heat absorbed by cold water
-9h=9c
- (mcAT)₁= (mcAT)c
whereAT=Tr-To
Transcribed Image Text:A. Heat Transfer and Temperature Change 1. Using the calorimetry simulator, determine the final temperature of a mixture of water when 110 grams of water at 90°C was added to 130 grams of water at 20°C. 2. Perform manual calculations to find the final temperature using the equation: (specific heat of water = 4.184J/g.°C) 3. Compare the results with the reading on simulator. 4. Calculate the amount of heat transferred in the process. heat given off by hot water heat absorbed by cold water -9h=9c - (mcAT)₁= (mcAT)c whereAT=Tr-To
A. Heat Transfer and Temperature Change
Mass of hot water (mp) = 110 g
Temperature of hot water (Th) =90 °C
Mass of cold water (mc) = 130 g
Temperature of cold water (Tc) = 20 °C
Final temperature of the system (T) from simulator = 52.08°C
Final temperature of the system (T) using manual calculations =
Amount of heat transferred (q): =
KJ
°C
Transcribed Image Text:A. Heat Transfer and Temperature Change Mass of hot water (mp) = 110 g Temperature of hot water (Th) =90 °C Mass of cold water (mc) = 130 g Temperature of cold water (Tc) = 20 °C Final temperature of the system (T) from simulator = 52.08°C Final temperature of the system (T) using manual calculations = Amount of heat transferred (q): = KJ °C
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