Question

A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 72°F room. After fifteen minutes, the internal temperature of the soup was 92°F To the nearest minute, how long will it take the soup to cool to 81°F? min

Answer 1

According to Newton's Law of cooling

The temperature after t minutes is given by

T(t) = T_s + (T_o - T_s)e^-kt

In this case, T_S = 72^oF

T_o = 100^oF

Therefore,

T(t) = 72 + (100 - 72)e^-kt

=> T(t) = 72 + 28e^-kt

Given, when t = 15 min, T(t) = 92^oF

Therefore,

92 =  72 + 28e^-k(15)

=> 28e^-15k = 92 - 72 = 20

=> e^-15k = 20 / 28 = 5 / 7

=> ln e^-15k = ln (5 / 7)

=> -15k = ln (5 / 7)

=> k = 0.022431

Therefore,

T(t) = 72 + 28e^{- 0.022431t}

Let us assume the temperature will be 81^oF after n minutes

Therefore,

72 + 28e^{- 0.022431n} = 81

=> 28e^{- 0.022431n} = 81 - 72 = 9

=> e^{- 0.022431n} = 9 / 28

=> ln e^{- 0.022431n} = ln (9 / 28)

=> - 0.022431n = ln (9 / 28)

=> n = 51 minutes

Therefore,

Answer: 51 min