# When a hot object is placed in a water bath whose temperature is 25 degree Celsius, it cools from...

## Question:

When a hot object is placed in a water bath whose temperature is 25 degree Celsius, it cools from 100 degree Celsius to 50 degree Celsius in 150 seconds. In another bath, the same cooling occurs in 130 seconds. Find the temperature of the second bath.

## Application of Newton's Law of Cooling:

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

{eq}\frac{dT}{dt} = - k(T - T_m) {/eq}

Where {eq}T_m {/eq} is the surroundings temperature and {eq}T {/eq} is the initial temperature of an object.

And its Solution is:

{eq}T(t) = T_m + ( T - T_m)e^{ -kt} {/eq}

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Accroding to the Newton's Law of Cooling we can write:

{eq}T(t) = 25 + ( T_1- 25 ) e^{-kt} {/eq}

Where, {eq}t {/eq} is the time measured in...