Question

The initial temperature of a cup of tea is 200ºF. The surrounding temperature is 70ºF, and the value of the constant k is 0.6. Applying Newton's cooling model, the temperature of the tea after 2 hours will be ºF

Answer 1

Newton's cooling model is ΔT = ΔTo * e ^ (-k t)

ΔTo = 200°F - 70°F = 130°F

k = 0.6

t = 2 hours

=> ΔT = 130 * e ^ (-0.6 t) = 130 * e^ (-0.6 * 2) = 130 * e ^ (-1.2)

ΔT = 39.15°F

ΔT = T - Tenvironment => T = ΔT + Tenvironment = 39.15°F + 70°F = 109.15°F ≈ 109 °F.

Answer: T = 109 °F