Question

The temperature in a hot tub is 104° and the room temperature is 75°. The water cools to 90° in 10 minutes. What is the water temperature after 20 minutes? (Round your answer to one decimal place.) ° After how many minutes will the temperature be 80°? (Round your answer to one decimal place.) min

Answer 1

Answer: (1) The temperature of the water after 20 minutes is 82.8°

(2) The temperature of the water will be 80° after 26.7 min

Step-by-step explanation: Please see the attachments below

Answer 2

   After 26.7 minutes, temperature of the the hot tub will be 80° .

Newton's cooling law:

• Newton's law of cooling is described by the expression,

        $T(t)=T_s+(T_0-T_s)e^{kt}$

        Here, $T(t)=$ Temperature of the object after 't' minutes

        $T_s=$ Temperature of surrounding

        $T_0=$ Initial temperature of the object

        $k=$ Constant for the given object

        $t=$ Time of cooling

Given in the question,

• Temperature of the hot tub $(T_0)=104^\circ$
• Room temperature $(T_s)=75^\circ$
• $T(t)=90^\circ$
• $t=10$ minutes

Substitute these values in the expression,

$90=75+(104-75)e^{10k}$

$15=29.e^{10k}$

$\text{ln}(\frac{15}{29})=\text{ln}(e^{10k})$

$-0.65925=10k$

$k=-0.065925$

Therefore, expression for the final temperature will be,

$T(t)=75+(104-75)e^{-0.065925t}$

$T(t)=75+(29)e^{-0.065925t}$

  Now substitute $T(t)=80^\circ$ to find the time of cooling in which temperature of the tub goes down to 80°C.

$80=75+(29)e^{-0.065925t}$

$5=(29)e^{-0.065925t}$

$e^{0.065925t}=\frac{29}{5}$

$\text{ln}(5.8)=0.065925t[\text{ln(e)}]$

$t=26.7\text{ minutes}$

      Hence, temperature of the the hot tub will be 80° after 26.7 minutes.

Learn more about the Newton's law of cooling here,

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