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. round to the nearest integer.

Newton's Law of Cooling states that the change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature over time.

Therefore when expressed mathematically, this is equivalent to:

dT = - k (T – Ts) dt

dT / (T – Ts) = - k dt

Integrating:

ln [(T2– Ts) / (T1– Ts)] = - k (t2 – t1)

Before we plug in the values, let us first convert the temperatures into absolute values R (rankine) by adding 460.

R = ˚F + 460

T1 = 200 + 460 = 660 R

Ts = 70 + 460 = 530 R

ln [(T2– 530) / (660 – 530)] = - 0.6 (2 - 0)

T2 = 569.16 R

T2 = 109 ºF

After 2 hours, it will be 109 ºF

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∠cbe and ∠deb are same-side interior angles.

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