Rate of Change Question4

In this page rate of change question4 we are going to see solution of some practice question of the worksheet.

Question 4:

Newton's law of cooling is given by θ = θ₀° e^(⁻kt), where the excess of temperature at zero time is θ₀° C and at time t seconds is θ° C. Determine the rate of change of temperature after 40 s ,given that θ₀ = 16° C and k = -0.03.(e^1.2 = 3.3201)

Solution:

Newton's law of cooling θ = θ₀° e^(⁻kt)

        θ₀° = 16° C

        k = -0.03

rate of change of temperature with respect to time

     dθ/dt = - k θ₀° e^(⁻kt)

t = 40

     dθ/dt = - (-0.03) (16)e^(⁻0.03) (40)

             =  0.48  e^(⁻1.2)

             =  0.48  (3.3201)

             =  1.5936° C/s