You are in Palm Spring, CA (elevation approx. 500 ft.) and the temperature is 100 degrees Fahrenheit. If you climb the mountain near there (Mt. San Jacinto), what will the approximate temperature be at the top (elevation approx. 10,500 ft) based on the average lapse rate?

Respuesta :

Answer:

The temperature is  [tex]T_t = 65 ^oF[/tex]

Step-by-step explanation:

From the question we are told that

   The elevation of the palm spring is  [tex]E_1 = 500 \ ft[/tex]

    The  temperature is  [tex]T = 100 ^oF[/tex]

    The elevation of  the mountain Mt. San Jacinto is  [tex]E_2 = 10500 ft[/tex]

Gnerally the difference in elevation between the land of the palm spring CA and the mountain is mathematically evaluated as

            [tex]\Delta E = E_2 - E_1[/tex]

=>        [tex]\Delta E = 10500 - 500[/tex]

=>         [tex]\Delta E = 10000 \ ft[/tex]

Gnerally the rate of decrease in temperature with increase in height is

          [tex]R = \frac{3.5 \ ^oF }{ 1000\ ft }[/tex]

Gnerally the decreases in temperature for the evaluated difference in elevation  is mathematically represented as

         [tex]D = \Delta E * R[/tex]

=>     [tex]D = 10000 * \frac{3.5}{1000}[/tex]

=>     [tex]D = 35^oF[/tex]

Gnerally the temperature at the top of the mountain  is

      [tex]T_t = T - D[/tex]

=>   [tex]T_t = 100 - 35[/tex]

=>   [tex]T_t = 65 ^oF[/tex]

 

   

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