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# Law of Cooling Calculator

Enter the required values and hit the calculate button to use Law of Cooling Calculator

## Law of Cooling Calculator

Law of Cooling Calculator is a tool that is used to find the temperature of an object in a specific time by using the initial temperature of an object, the surrounding’s constant temperature of the body, and the passing time to cool according to its surroundings.

## What is the law of Cooling?

The Law of Cooling tells us how the object’s temperature changes with time when it contact with its surrounding while the object cannot get the temperature continuously. The law of cooling is defined as a change in the temperature that is proportional to the temperature of the surroundings and the difference in the temperature of the object from initial to ambient temperature.

Its mathematical formula is stated as.

**T(t) = T _{a} + (T_{o} – T_{a}) **

**×**

**e**

^{(-kt)}Where

- T (t) = Temperature of the object or Core Temperature
- T
_{a}= Surrounding constant temperature - T
_{o}= Initial temperature of the object - K = Cooling constant
- t = cooling time taken about its surroundings

## How to Find the Core Temperature?

**Example 1:**

Evaluate the temperature of the hot cup of tea after 10 minutes by using the law of cooling while the temperature of the is surrounding is 25^{0} C and the initial temperature of the hot cup of tea is 90^{0} C with the core temperature of the cup being 60^{0} C.

**Solution:**

**Step 1:**

**Write the data from the above conditions carefully.**

T (t) = Temperature of the object or Core Temperature = 60,

T_{a} = Surrounding constant temperature = 25^{0} C

T_{o} = Initial temperature of the object = 90^{0} C

K = Cooling constant =?, t = cooling time taken = 10 min, T (t) = Core Temperature =?

**Step 2:**

**Write the Formula of the Law of Cooling in detail.**

T(t) = T_{a} + (T_{o} – T_{a}) × e^{(-kt)}

**Step 3:**

**To find the core temperature first we find the value of the Cooling Constant by putting the values in the above formula.**

T (t) = 60^{0} C, T_{a} = 25^{0} C, T_{o} = 90^{0} C, t = 10 min, K = Cooling constant =?

T (t) = T_{a} + (T_{o} – T_{a}) × e^{(-kt)}

60 = 25 + (90 – 25) × e^{(-k(10))}

60 – 25 = (90 – 25) × e^{(-k(10))}

(60 – 25)/(90 – 25) = e^{(-k(10))}

e^{(-k(10))} = (60 – 25)/(90 – 25)

e^{(-k(10))} = 35/65

Apply the natural logarithm on both sides and simplify.

ln [e^{(-k(10))}] = ln [35/65]

-10k = ln [35/65]

-10k = - 0.6190

k = (- 0.6190)/(-10)

**k = 0.0619**

**Step 4:**

**Put the above “K” value and other data carefully in the “Step 2” formula.**

T (t) =?, T_{a} = 25^{0} C, T_{o} = 90^{0} C, t = 10 min, K = 0.0619

T(t) = T_{a} + (T_{o} – T_{a}) × e^{(-kt)}

T(t) = 25 + (90 – 25) × e^{[-(0.0619)(10)]}

T(t) = 25 + (65) × e^{[-(0.619)]}

T(t) = 25 + (65) × (0.5385)

T(t) = 25 + 35

**T(t) = 60 ^{0} C**