Newton's Law of Cooling Calculator

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**Step 1 :**

**Step 2 :**

Newton’s Law of Cooling can be define as “the rate of change of the temperature of an object is proportional to the difference in temperatures between the body and its surroundings.”

**Newton's Law of Cooling Calculator** will help to calculate the temperature of the object at time(t).

Observe the value of surrounding constant temperature(T_{s}), Initial temperature of the object(T_{o}), core temperature(T) and time(t).

Apply the formula:

Temperature of the object at time t, T (t) = T_{s} + (T_{o} - T_{s})*e^{(-k*t)}

Calculate law of cooling of the object for the given details

Surrounding constant temperature (Ts) = 40

Initial temperature of the object (To) = 20

Core Temperature (T) = 30

Time(t) = 22**Step 1 :**Given that:

Surrounding constant temperature (Ts) = 40

Initial temperature of the object (To) = 20

Core Temperature (T) = 30

Time(t) = 22**Step 2 :**Temperature of the object at time t, T (t) = T

_{s}+ (T_{o}- T_{s})*e^{(-k*t) }

k = log($\frac{(30-40)}{(20-40)}$) = -0.301029996

T (t) = 40 + (20 - 40)*e

^{(-0.301029996*22) }

T (t) = 39.973 F

**Answer :**T (t) = 39.973 F

Calculate law of cooling of the object for the given details

Surrounding constant temperature (Ts) = 50

Initial temperature of the object (To) = 20

Core Temperature (T) = 20

Time(t) = 22**Step 1 :**Given that:

Surrounding constant temperature (Ts) = 50

Initial temperature of the object (To) = 20

Core Temperature (T) = 20

Time(t) = 22**Step 2 :**Temperature of the object at time t, T (t) = T

_{s}+ (T_{o}- T_{s})*e^{(-k*t) }

k = log($\frac{(20-50)}{(20-50)}$) = 0

T (t) = 50 + (20 - 50)*e

^{(-0*22) }= 50 + (-30) = 50 - 30 = 20

T (t) = 20 F

**Answer :**T (t) = 20 F