
Yefim S. answered 12/22/21
Math Tutor with Experience
T(t) = Ts + (To – Ts) e-kt; T(10) = 60 = 20 + (90 - 20)e-10k; 40 = 70e-10k; k = - 0.1ln(4/7) = 0.056 min-1
Lemigio K.
asked 12/22/21A hot cup of coffee at 90°C is placed on a table in the kitchen where the room temperature is 20°C. Using Newton's law of cooling, find at what rate the temperature of the coffee was decreasing when it was placed on the table. Give correct units.
Additional information: After 10 minutes the temperature of the coffee was 60°C.
Yefim S. answered 12/22/21
Math Tutor with Experience
T(t) = Ts + (To – Ts) e-kt; T(10) = 60 = 20 + (90 - 20)e-10k; 40 = 70e-10k; k = - 0.1ln(4/7) = 0.056 min-1
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Lemigio K.
Isn't -0.0556 the value for k? Now T = 20 + 70e^-kt. Don't we take the derivative of the general solution and plug in 0 for t since we are looking for the rate of change of temperature w.r.t.time?12/22/21