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A Warming Liquid - A liquid is taken out of a refrigerator and placed in a warmer room, where its temperature, in F, increases over time. It can be modeled using the equation Tm 74 390.87m. (a) What temperature did the liquid start at? Show the work that leads to your answer. (b) What is the temperature of the room?

Sagot :

MrRoyal

Answer:

The temperature at the start is 35F

The temperature of the room is 74F

Step-by-step explanation:

Given

[tex]T(m) = 74 - 39(0.87)^m[/tex]

Solving (a): The temperature at the start.

To do this, we set: [tex]m = 0[/tex]

So, we have:

[tex]T(0) = 74 - 39(0.87)^0[/tex]

[tex]T(0) = 74 - 39*1[/tex]

[tex]T(0) = 74 - 39[/tex]

[tex]T(0) = 35[/tex]

Hence, the temperature at the start is 35F

Solving (b): The room temperature

We have:

[tex]T(m) = 74 - 39(0.87)^m[/tex]

To get the temperature of the room, we simply remove the exponential function.

So, we have:

[tex]Initial = 74[/tex]

Hence, the temperature of the room is 74F

The initial temperature of the liquid is 35 F and the room temperature is 74 F.

What is Melting Point?

The melting point is the temperature at which the solid starts converting into liquid.

Given here,

[tex]T_m = 74- 39(0.87)^m[/tex]

Leat the [tex]m = 0,[/tex]

So,

[tex]T_0 = 74- 39(0.87)^0\\\\T_0 = 35 \rm \ F[/tex]

Thus the initial temperature is 35 F.

Now remove the exponential to get the room temperature,

So,

[tex]T_{RT} = 74[/tex]

Therefore, the initial temperature of the liquid is 35 F and the room temperature is 74 F.

Learn more about the Melting point:

https://brainly.in/question/608391

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