By newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. if a certain object cools from 115° to 90° in half an hour when surrounded by air at 70°, find its temperature at the end of another half hour.

The temperature at the end of another half hour is 78.85°.

From Newton's law of cooling;

T(t) = T(s) + (To - Ts)e^kt

Where;

T(t) = temperature at time t = 90°

T(s) = temperature of the surrounding = 70°

To = initial temperature = 115°

k = cooling constant

t = time taken = 30 mins

Substituting values;

90 = 70 + (115 - 70)e^30k

90 - 70 =  (115 - 70)e^30k

20 = 45e^30k

20/45 = e^30k

0.444 = e^30k

ln0.444 = lne^30k

ln0.444 = 30k

k = ln0.444/30

k = -0.0271

At the end of another half hour, 60 minutes have now gone.

T(t) = 70 +  (115 - 70)e^(-0.0271)(60)

T(t) = 78.85°

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Explanation:

As per Newton's Law of cooling we know that

now we have

temperature cool from 115 to 90 degree C in half an hour

so we have

now we will say in next half an hour temperature changes to T

it about 600 mph from what i calculated

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures