Question
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
- 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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