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A student in a science lab is investigating heat transfer and thermal energy conservation as she mixes hot and cold water. She measures out her

A student in a science lab is investigating heat transfer and thermal energy conservation as she mixes hot and cold water. She measures out her desired amount of cold water and hot water into different Styrofoam cups. She measures the temperatures of each before pouring the hot water into the cold water. She monitors the temperature of the mixed waters and records the final temperature. She does three trials, which are described below.

  • Trial 1: She mixed 250 mL of water at 20 C with 250 mL of water at 98 C. After waiting some time, she recorded the temperature of the mix to be 56 C.
  • Trial 2: She mixed 200 mL of water at 20 C with 400 mL of water at 98 C. After waiting some time, she recorded the temperature of the mix to be 72 C.
  • Trial 3: She mixed 300 mL of water at 15 C with 150 mL of water at 90 C. After waiting some time, she recorded the temperature of the mix to be 41 C.

Data Table


Trial Volume of Cold Water (mL) Temperature of Cold Water (oC) Volume of Hot Water (mL) Temperature of Hot Water (oC) Temperature of Mixture (oC)
1 250mL 20oC 250mL 98oC 56oC
2 200mL 20oC 400mL 98oC 72oC
3 300mL 15oC 150mL 90oC 41oC

Data Calculations

Trial Mass of Cold Water (kg) Change in Temperature of Cold Water (oC) Mass of Hot Water (kg) Change in Temperature of Hot Water (oC)
1 0.25kg 5620=36C 0.25kg 5698=42C
2 0.2kg 7220=52C 0.4kg 7298=26C
3 0.3kg 4115=26C 0.15kg 4190=49C

Questions

  1. Calculate the heat gained, Q, by the cold water foreach trial. Show your work.

Trial 1:

Equation: Q = (m)(c)(T) (0.25kg)(4.186J/gC)(36C) = 37.395kJ

Trial 2:

Equation: Q = (m)(c)(T)

(0.20kg)(4.186J/gC)(52C) = 43.676kJ

Trial 3:

Equation: Q = (m)(c)(T)

(0.30kg)(4.186J/gC)(26C) = 32.818kJ

2. Calculate the heat lost, Q, of the hot water foreach trial. Show your work.

Trial 1:

Equation: Q = (m)(c)(T)

(0.25kg)(4.186J/gC)(42C) = 43.953kJ

Trial 2:

Equation: Q = (m)(c)(T)

(0.40kg)(4.186J/gC)(26C) = 43.676kJ

Trial 3:

Equation: Q = (m)(c)(T)

(0.15kg)(4.186J/gC)(49C) = 31.155kJ

3. Compare the values for temperature changes and heat gain and heat loss in the previous questions for each trial.

 

a. Describe what you notice. ( in your own words )

 

 

 b. Why do you think this occurred? ( in your own words)

 

4. In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. Now look at your answer to question 3. What might have caused the difference you have reported? (Even though this data was provided to you, think of the errors the student could have encountered when collecting the data.)

5. Write, a complete conclusion for this activity. (Remember to include the purpose, summary of results, and possible sources of error)


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