BRAVO! That is correct.

A cup of hot tea, initially at 95^oC, cools to 75^oC, in 5.0 minutes when sitting in 25^oC sirrpimdomgs. Use Newton's Law of Cooling to determine how long it will take to cool to 35^oC.

Newton's Law of Cooling is

T_surf is T_surf(t), the temperature of the surface as a function of time t.

T_o is the temperature of the surroundings, T_o = 25^oC .

T is the initial temperature difference; T = 95^oC - 25^oC = 70 C^o.

We know T_surf at time t = 5.0 min; that is 75^oC.

T_surf = T_o + T e ^{- t/}

75^oC = 25^oC + (70^oC)e ^{- 5 min/}

We know everything in this equation except the time constant . This describes the cooling rate and is determined by the physical situation. (Solving for gives opportunity to play with the natural logrithm and exponential functions on a calculator).

(70^oC)e ^{- 5 min/} = 75^oC - 25^oC = 50^oC

e ^{- 5 min/} = 50^oC/70^oC

e ^{- 5 min/} = 0.714

How can we get out of this exponent? Recall -- from a long-ago algebra class -- that ln(e^x) = x. That means that

ln(e ^{- 5 min/} ) = - 5 min/

So we take the ln() -- the "natural logrithm" function -- of both sides of the equation,

This is the time constant for the exponent in Newton's Law of Cooling. Now we can evaluate T_surf for any time t. Or we can calculate the time t for any temperature T_surf. What time t gives T_surf = 35^oC?

T_surf = T_o + T e ^{- t/}

35^oC = 25^oC + (70^oC)e ^{- t/14.84 min}

35^oC - 25^oC = (70^oC)e ^{- t/14.84 min}

10^oC = (70^oC)e ^{- t/14.84 min}

(70^oC)e ^{- t/14.84 min} = 10^oC

e ^{- t/14.84 min} = 10^oC/70^oC

e ^{- t/14.84 min} = 0.143

And now -- just as before -- we need to use the ln() -- the "natural logrithm" function -- to get the time t out of the exponent,