Ask Question
25 February, 08:59

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)
  1. 25 February, 11:24
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Newton's law of cooling states that for a cooling substance with initial temperature T0, the temperature T (t) after t minutes can be ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers