NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Newton's law of cooling

## Interactive practice questions

A warm object is placed in a cooler room of constant temperature and cools at a rate proportional to the difference between its temperature and the room’s temperature.

a

When is the difference in temperature between the object and the room the greatest?

As $t$t approaches infinity.

A

When the object is initially put in the room.

B

After $10$10 minutes.

C

As $t$t approaches infinity.

A

When the object is initially put in the room.

B

After $10$10 minutes.

C
b

So when will the object be decreasing in temperature most rapidly?

When it is initially put in the room.

A

After the first minute.

B

As $t$t approaches infinity.

C

When it is initially put in the room.

A

After the first minute.

B

As $t$t approaches infinity.

C
c

Which of the following statements are true? Select all that apply.

As the object cools, the difference in temperature decreases so the more time that passes, the more quickly the object reaches the room temperature.

A

As the object cools, the difference in temperature decreases so the rate of cooling increases.

B

As the object cools, the difference in temperature decreases so the object cools down more slowly over time.

C

As the object cools, the difference in temperature decreases so the rate of cooling slows down.

D

As the object cools, the difference in temperature decreases so the more time that passes, the more quickly the object reaches the room temperature.

A

As the object cools, the difference in temperature decreases so the rate of cooling increases.

B

As the object cools, the difference in temperature decreases so the object cools down more slowly over time.

C

As the object cools, the difference in temperature decreases so the rate of cooling slows down.

D
d

Which of the following shows how the temperature of a warm object changes over time when it is placed in a room of constant temperature $28$28$°C$°C?

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A

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B

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C

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D

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A

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B

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C

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D
Easy
Approx 2 minutes
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When a glass vase is made, the glass reaches a temperature of $1010$1010$°C$°C (degrees celsius) and is then placed in a cooling room where the temperature is a constant $15$15$°C$°C. The rate at which the glass vase cools is given by $\frac{dT}{dt}=-0.019\left(T-15\right)$dTdt=0.019(T15), where $T$T is the temperature of the vase $t$t minutes after being placed in the room.

When an object is heated to a certain temperature and then placed in an environment of constant temperature $25$25$°C$°C, it starts to cool and the rate at which it cools is given by $\frac{dT}{dt}=-0.002\left(T-25\right)$dTdt=0.002(T25).

The temperature of the object $T$T after $t$t minutes of cooling is given by $T=25+1010e^{-0.002t}$T=25+1010e0.002t.

Consider two metals $M$M and $P$P that are heated to a temperature of $1050$1050$°C$°C. They are both placed in the same room where the temperature is a constant $25$25$°C$°C, so that the rate of cooling for each metal can be given by $\frac{dT}{dt}=-k\left(T-25\right)$dTdt=k(T25) and the model for temperature $T$T at time $t$t can be given by $T=25+1025e^{-kt}$T=25+1025ekt.

The temperature of each metal over time is graphed below.

### Outcomes

#### M8-12

Form differential equations and interpret the solutions

#### 91578

Apply differentiation methods in solving problems