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The food starts at 25^ extoC and cools at 6^ extoC per hour until it reaches the freezer temperature of -8^ extoC. What is the temperature of the food after 10 hours and after 7 hours?

Mathematics

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QUESTION 4

The food starts at 25∘C25^\circ C and cools at 6∘C6^\circ C per hour until it reaches the freezer temperature of −8∘C-8^\circ C.

First, calculate the time it takes for the food to reach the freezer temperature:

  1. Calculate the total temperature drop required to reach −8∘C-8^\circ C.

    Temperature drop=25∘C−(−8∘C)=25∘C+8∘C=33∘C\text{Temperature drop} = 25^\circ C - (-8^\circ C) = 25^\circ C + 8^\circ C = 33^\circ C
  2. Calculate the time needed to achieve this drop.

    Time to reach freezer temp=33∘C6∘C/hour=5.5 hours\text{Time to reach freezer temp} = \frac{33^\circ C}{6^\circ C/\text{hour}} = 5.5 \text{ hours}

    Since the food cannot cool below the freezer temperature of −8∘C-8^\circ C, if the given time is greater than 5.5 hours, the temperature will be −8∘C-8^\circ C.

    a) 10 hours

  3. Compare the given time to the time needed to reach freezer temperature.

    Since 10 hours>5.5 hours10 \text{ hours} > 5.5 \text{ hours}, the food will have already reached the freezer temperature.
    The temperature of the food after 10 hours will be −8∘C\boxed{\mathbf{-8^\circ C}}.

    b) 7 hours

  4. Compare the given time to the time needed to reach freezer temperature.

    Since 7 hours>5.5 hours7 \text{ hours} > 5.5 \text{ hours}, the food will have already reached the freezer temperature.
    The temperature of the food after 7 hours will be −8∘C\boxed{\mathbf{-8^\circ C}}.

    QUESTION 5

    a) After a dry season, the river level is −3-3 metres. How many metres would the river need to rise to reach flood level?

  5. Assume flood level is 0 metres.

    Calculate the difference.

    Rise needed=0 m−(−3 m)=0+3=3 m\text{Rise needed} = 0 \text{ m} - (-3 \text{ m}) = 0 + 3 = \boxed{\mathbf{3 \text{ m}}}

    b) If the level rose from −3-3 m to −2-2 m, how much did it rise?

  6. Calculate the difference between the final and initial levels.

    Rise=−2 m−(−3 m)=−2+3=1 m\text{Rise} = -2 \text{ m} - (-3 \text{ m}) = -2 + 3 = \boxed{\mathbf{1 \text{ m}}}

    c) In the rainy season, the level rose from −2-2 m to 1.5 m. By how much did it rise?

  7. Calculate the difference between the final and initial levels.

    Rise=1.5 m−(−2 m)=1.5+2=3.5 m\text{Rise} = 1.5 \text{ m} - (-2 \text{ m}) = 1.5 + 2 = \boxed{\mathbf{3.5 \text{ m}}}

    d) When some of the water had drained away, the depth fell from 1.5 m to −0.6-0.6 m. By how many metres had it fallen?

  8. Calculate the difference between the initial and final levels to find the fall.

    Fall=1.5 m−(−0.6 m)=1.5+0.6=2.1 m\text{Fall} = 1.5 \text{ m} - (-0.6 \text{ m}) = 1.5 + 0.6 = \boxed{\mathbf{2.1 \text{ m}}}
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