Question

When a hot object is placed in a water bath whose temperature is 25◦C, it cools from 100 to 50◦C in 150 s. In another bath, the same cooling occurs in 120 s. Find the temperature of the second bath.

Answer 1

Using Newton's Law of Cooling, the given are:
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s

dT/dt=-k(T-Ta)
$\int\limits^a_ b{(dT/(T-Ta1))} \, dt = \int\limits^x_y{-k} \, dt$
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150

Solving for k=7.324081X10^-3

For the second bath: find Ta2 with dt=120s

dT/dt=-(7.324081x10^-3)*(T-Ta2)
$\int\limits^a_ b{(dT/(T-Ta2))} \, dt = \int\limits^x_y{-7.324081x10^-3} \, dt$
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(7.324081x10^-3)*120

Solving for Ta (second bath temperature)= 14.49 degrees Celsius

Answer 2

Answer:

the temperature of the second bath is 20.41 C.

Explanation