Respuesta :

skyluke89

Answer:

Here we want to calculate the sum of the temperatures in each beaker at each time.

To do that, we just add the values of the temperatures in each beaker for each time.

At time t = 0 s,

[tex]\sum T = 95+5+90=190^{\circ}C[/tex]

At time t = 100 s,

[tex]\sum T = 77+23+54=154^{\circ}C[/tex]

At time t = 200 s,

[tex]\sum T = 67+33+34=134^{\circ}C[/tex]

At time t = 300 s,

[tex]\sum T = 60+40+20=120^{\circ}C[/tex]

At time t = 400 s,

[tex]\sum T = 56+44+12 = 112^{\circ}C[/tex]

At time t = 500 s,

[tex]\sum T = 54+46+8=108^{\circ}C[/tex]

At time t = 600 s,

[tex]\sum T = 52+48+4=104^{\circ}C[/tex]

The sum of the temperatures in each beaker is calculated below.

We want to find the sum of the temperatures in each beaker at each time;

At t = 0 s; ΣT = 95 °C + 5 ° C = 100

At t = 100 s; ΣT = 77 °C + 23 ° C = 100

At t = 200 s; ΣT = 67 °C + 33 ° C = 100

At t = 300 s; ΣT = 60 °C + 40 ° C = 100

At t = 400 s; ΣT = 56 °C + 44 ° C = 100

At t = 500 s; ΣT = 54 °C + 46 ° C = 100

At t = 600 s; ΣT = 52 °C + 48 ° C = 100

Read more at; https://brainly.com/question/18482560

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