Newton's law of cooling states that for a cooling substance with initial temperature T0, the temperature T (t) after t minutes can be modeled by the equation T (t) = Ts + (T0-Ts) e-kt, where Ts is the surrounding temperature and k is the substance's cooling rate.

A liquid substance is heated to 80°C. Upon removing it from the heat it cools to 60°C in 12 minutes. What is the substance's cooling rate when the surrounding air temperature is 50°C?

Round the answer to four decimal places.

Question

Jaylene Alvarado

Mathematics

8 February, 01:09

Newton's law of cooling states that for a cooling substance with initial temperature T0, the temperature T (t) after t minutes can be modeled by the equation T (t) = Ts + (T0-Ts) e-kt, where Ts is the surrounding temperature and k is the substance's cooling rate.

A liquid substance is heated to 80°C. Upon removing it from the heat it cools to 60°C in 12 minutes. What is the substance's cooling rate when the surrounding air temperature is 50°C?

Round the answer to four decimal places.

+3

Answers (1)

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Melanie Livingston · Answer 1

T (t) = Ts + (T0 - Ts) e^-kt

60 = 50 + (80 - 50) e^-12k

30e^-12k = 60 - 50 = 10

e^-12k = 10/30 = 1/3

-12k = ln (1/3)

k = ln (1/3) / - 12 = 0.0916

k = 0.0916