Unformatted text preview: (1 point)
When a hot object is placed in a water bath whose temperature is 25"C, it cools from 100"C to 50"C in 195s. In another bath, the same
cooling occurs in 175s. Find the temperature of the second bath.
The temperature of the second bath = 20.1025
Solution: With To = 25'C, the temperature of the object is given by
F(t) = 25 + Ce * for some constants C and k. From the initial condition,
F(0) = 25 + C = 100, so C = 75. After 195 seconds,
F(195) = 25 + 75e-195k = 50, so
k =--osln 2% ~ 0.00563391s-1
If we place the same object with a temperature of 100"C into a second bath whose temperature is To , then the temperature of the
object is given by
F(t) = To + Ce-0.0056339It, where F(0) = To + C = 100, so
F(t) = To + (100 - To)e-0.00563391t
To cool from 100'C to 50"C in 175s, To must satisfy
To + (100 - To)e-0.00563391(175) = 50.
Thus, To = 20.2437'C....
View Full Document
- Fall '14
- Harshad number, e-0.00563391t