Chemistry/Measuring Temperature Marking Scheme

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Marking Scheme Science 8 Laboratory Investigation
MEASURING TEMPERATURE

Maximum = 9 + 2 (Bonus)

Data Step Volume used Temperature
1 (b) 100 mL of cold water 14
100 mL of hot water 50
1 (c) 200 mL of mixture 32
2 (b) 50 mL of cold water 22
100 mL of hot water 52
2 (c) 150 mL of mixture 42
3 (b) 100 mL of cold water 15
50 mL of hot water 51
3 (c) 150 mL of mixture 27

(1)

Questions
1. This question uses the data for 100 mL of hot water and 100 mL of cold water.
(a) Calculate the starting temperature difference between the hot water and cold water, using the data for step 1 (b).

(1/2) Starting temperature difference = 50 – 14 = 36oC

(b) How many degrees does the temperature rise when the hot water is added to the cold water (to create 200 mL of mixture)?

(1/2) Temperature rise = 32 – 14 = 18oC

(c) Calculate the percentage of the mixture which is hot water as follows.


Percentage of hot water = [math]\frac {Volume of hot water} {Volume of mixture}[/math] * 100%


(1) Percentage of hot water = [math]\frac {100mL} {200mL}[/math] * 100% = 50%


(d) Calculate the percentage temperature rise when hot water is added as follows.

Percentage temperature rise = [math]\frac {Temperature rise} {Starting Temperature Difference}[/math] *100%

(1) Percentage temperature rise = [math]\frac {18^\circ C} {36^\circ C}[/math] *100% = 50%


2. Repeat all the calculations for question 1, using the data for 100 mL of hot water and 50 mL of cold water.
(1)

(a) Starting temperature difference = 52 – 22 = 30oC
(b) Temperature rise = 42 – 22 = 20oC
(c) Percentage of hot water = [math]\frac {100mL} {150mL} [/math] * 100% = 67%
(d) Percentage temperature rise = [math]\frac {20^\circ C} {30^\circ C} [/math] * 100% = 67%


3. Repeat all the calculations for question 1, using the data for 50 mL of hot water and 100 mL of cold water.
(1)

(a) Starting temperature difference = 51 – 15 = 36oC
(b) Temperature rise = 27 – 15 = 12oC
(c) Percentage of hot water = [math]\frac {50mL} {150mL}[/math] * 100% = 33%
(d) Percentage temperature rise = [math]\frac {12^\circ C} {36^\circ C}[/math] * 100% = 33%


4. Copy the results of your calculations for questions 1, 2 and 3 into the following table.
(1)

Mixture Percentage of hot water Percentage temperature rise
100 mL hot + 100 mL cold 50 50
100 mL hot + 50 mL cold 67 67
50 mL hot + 100 mL cold 33 33

5. Lets make sure we know what the entries in the table mean.

  • “Percentage of hot water” tells us what % of a mixture is made up of hot water.
  • “Percentage temperature rise” tells us what the temperature rise is, as a percentage of the starting temperature difference.

Write a sentence or statement that describes any similarities you observe in the table for question 4.

(1) When hot and cold water is mixed, the percentage temperature rise of the cold water is equal to the percentage of hot water added.


7. If you mix 500 mL of cold water with 700 mL of hot water, will you expect the temperature of the mixture to be closer to the starting temperature of the cold water or closer to the starting temperature of the hot water?

(1) Because there is more hot water than cold water, the temperature of the mixture will be closer to the temperature of the hot water than to the temperature of the cold water.


8. BONUS QUESTION: If you mix 120 mL of water at 90oC with 30 mL of water at 15oC, what is the final temperature of the mixture?
(2)

Starting temperature difference = 90 – 15 = 75oC

Percentage of hot water = [math]\frac {120mL} {150mL} [/math] * 100% = 80%
Temperature will rise by 80% of 75oC = 0.80 x 75 = 60oC
Final temperature = 15 + 60 = 75oC