Question

A pot of boiling soup with an internal temperature of 100 degree F was taken off the solve to cool in a 69 degree F room. After 15 mintues the internal temperature of the soup was 95 degree F. To the nearest minute, how long will it take the soup to cool to 80degree F?

Answer 1

Answer:

86.34 min

Step-by-step explanation:

Using the formula for Newton's law of cooling.

$T = T_{1} + (T_{0} - T_{1})e^{kt}$

where T₀ = initial temperature of object = 100 F

T₁ = temperature of surrounding = 69 F

T = final temperature of object and

t = time

Initially, when the temperature of the soup drops to 95 F, it takes 15 minutes. So, T = 95 F and t = 15 min

Substituting all the variables into the equation for T we have

$95 = 69 + (100 - 69)e^{15k}\\95 - 69 = 31e^{15k}\\26 = 31e^{15k}\\e^{15k} = 26/31\\e^{15k} = 0.8387\\15k = ln(0.8387)\\k = ln(0.8387)/15\\k = -0.1759/15\\k = -0.012$

So,

$T = 69 + (100 - 69)e^{-0.012t}\\T = 69 + 31e^{-0.012t}$

When T = 80 F,

$80 = 69 + 31e^{-0.012t}\\ 80 - 69 = 31e^{-0.012t}\\11 = 31e^{-0.012t}\\e^{-0.012t} = 11/31\\e^{-0.012t} = 0.3548\\-0.012t = ln(0.3548)\\t = -ln(0.3548)/0.012\\t = 1.0361/0.012\\t = 86.34$

So, it takes the soup 86.34 minutes to cool to 80 °F