# Chemistry/Measuring Temperature Marking Scheme

**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 = **36 ^{o}C**

(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 = **18 ^{o}C**

(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 =
**30**^{o}C - (b) Temperature rise = 42 – 22 =
**20**^{o}C - (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**%

- (a) Starting temperature difference = 52 – 22 =

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 =
**36**^{o}C - (b) Temperature rise = 27 – 15 =
**12**^{o}C - (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%**

- (a) Starting temperature difference = 51 – 15 =

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 90^{o}C with 30 mL of water at 15^{o}C, what is the final temperature of the mixture?

(2)

- Starting temperature difference = 90 – 15 = 75
^{o}C

- Starting temperature difference = 90 – 15 = 75

Percentage of hot water = [math]\frac {120mL} {150mL} [/math] * 100% = 80%

Temperature will rise by 80% of 75^{o}C = 0.80 x 75 = 60^{o}C

Final temperature = 15 + 60 = **75 ^{o}C**