Question
The two insulated beakers contain equal amounts of identical liquids. The temperature of Beaker A is 85°C. The temperature of Beaker B is 40°C. A metal rod connects the beakers. Five minutes later, the temperature of Beaker A is 75°C and the temperature of Beaker B is 50°C. What might the temperatures be after another five minutes have passed?
A. Temp. of Beaker A = 68°C, Temp. of Beaker B = 57°C
B. Temp. of Beaker A = 64°C, Temp. of Beaker B = 61°C
C. Temp. of Beaker A = 68°C, Temp. of Beaker B = 48°C
D. Temp. of Beaker A = 68°C, Temp. of Beaker B = 53°C
i think the answer is b
A. Temp. of Beaker A = 68°C, Temp. of Beaker B = 57°C
B. Temp. of Beaker A = 64°C, Temp. of Beaker B = 61°C
C. Temp. of Beaker A = 68°C, Temp. of Beaker B = 48°C
D. Temp. of Beaker A = 68°C, Temp. of Beaker B = 53°C
i think the answer is b
Answers
bobpursley
The temperature changes are exponential.
The initial difference was 68-40=28
after 5 min, the difference was 25 C
after infinity, the difference will be 0C
so I model the temp differnece as
TempDiff=28e^kt
so lets solve for k
at the first five min mark
25=28e^5k
or ln (25/28)=5k
and we solve for k, I get
k=-.022666 or close to that, check it.
so at the next five min...
temp diff=28e^k(10)=22.32
Answer A is the closest
There are a number of other ways to think this out.
The initial difference was 68-40=28
after 5 min, the difference was 25 C
after infinity, the difference will be 0C
so I model the temp differnece as
TempDiff=28e^kt
so lets solve for k
at the first five min mark
25=28e^5k
or ln (25/28)=5k
and we solve for k, I get
k=-.022666 or close to that, check it.
so at the next five min...
temp diff=28e^k(10)=22.32
Answer A is the closest
There are a number of other ways to think this out.
i don't really understand your work. i'm oly in 6th grade can u make it a little easier
bobpursley
sixth grade, ok, I figured you were in college. This is a college question. give me a couple of minutes.
bobpursley
Ok, sixth grader answer. You know the final temp in each will be an equal 62.5C (half way between 85 and 40). Think out why..there were equal amounts of liquid, the same liquid. So each has to change by half way, or 22.5/2 = 11.2C.
Now look at liquid A. It stated at 85, in five minutes it went down 10 deg, or 10/22.5 or almost half way to the final. You know in the next five min, it is likely to go about near half the rest of the way..rest of the way is 75-62.5=13.5, so half that is 6.7C
so liquid A is going to be near 85-10-6.7 or near 68C or 5.5C from the final 62.5C temp.
Now using the same thinking, liquid B should be about 5.5C below 65 final temp, or at 59C
Answer A is the best answer.
Now look at liquid A. It stated at 85, in five minutes it went down 10 deg, or 10/22.5 or almost half way to the final. You know in the next five min, it is likely to go about near half the rest of the way..rest of the way is 75-62.5=13.5, so half that is 6.7C
so liquid A is going to be near 85-10-6.7 or near 68C or 5.5C from the final 62.5C temp.
Now using the same thinking, liquid B should be about 5.5C below 65 final temp, or at 59C
Answer A is the best answer.
where did u get the 62.5 degrees celsius from?
bobpursley
initial temp difference is 85-40=45 C
half way then is 85-45/2=62.5 C
half way then is 85-45/2=62.5 C
ok thanks