Two large containers A and B of the same size are lled with different uids. The uids in containers A and B are maintained at 0° C and 100° C, respectively. A small metal bar, whose initial temperature is 100° C, is lowered into container A. After 1 minute the temperature of the bar is 90° C. After 2 minutes the bar is removed and instantly transferred to the other container. After 1 minute in container B the temperature of the bar rises 10°. How long, measured from the start of the entire process, will it take the bar to reach 99.9° C?

t = 9.05 min

Step-by-step explanation: The rate of temperature over time is given by:

where:

K is the constant of transference of heat

is temperature of the container

ln (T - ) = kt + c

In container A, the temperature is at 0°C, so

For the bar, when t = 0 min, T = 100°C:

100 =

C = 100

After 1 minute, the temperature of the bar is 90°C, so:

k = ln(0.9)

k = - 0.105

After 2 minutes, the temperature will be:

T = 81.06°C

For container B, the temperature is 100°C, so:

The initial temperature of the bar when entering the container B is T = 81.06°C, then:

C = - 18.94

After 1 minute, the temperature rises 10°C:

91 =

k = ln(0.475)

k = - 0.744

When T = 99.9°C:

t =

t = 7.05 min

The entire process will take:

t = 2 + 7.05

t = 9.05 min

answer:

step-by-step explanation: