Temperature change- algebra 1

Question

The temperature changes of two calorimeters were monitored and recorded Calorimeter A contained 250 milliliters of water with an initial water temperature of 120°F After 10 minutes, the temperature of the water in Calorimeter A steadily decreased to 100°F Calorimeter B contained the same amount of water as Calorimeter A but its initial water temperature was 20°F The rate of the increase in water temperature in Calorimeter B was 3°F per minute.

A) Write a system of linear equations for the temperature, T °F, of the

water in Calorimeters A and B in terms of the time, t minutes.

B) After how many minutes will the temperature of the water in the two

calorimeters be the same? How do you know?

Answer 1

A)

T_Ao = 120°F is the initial temperature of calorimeter A

T_Bo = 20°F is the initial temperature of calorimeter B,

r_A = unkown is the rate of change of the temperature in calorimeter B

r_B = 3°F/min is the rate of change of the temperature in calorimeter B

Since calorimeter A decreases 20°F in 10 minutes, this gives r_A = -20°F /10 min = -2°F/min. Its negative because the temperature is decreasing

For "t" the minutes since the start, we get the following system of equations:

T_A = r_A * t + T_Ao = -2°F/min * t + 120°F gives the temperature of A at any time

T_B = r_B * t + T_Bo = 3°F/min * t + 20°F gives the temperature of B at any time

B)

To find we the temperatures are the same, we want

T_A = T_B or

-2°F/min * t + 120°F = 3°F/min * t + 20°F which gives

100°F = 5°F/min * t so

t = 100°F / 5°F/min = 20 min

Let's check. Substitute 20 min in both equations and we get T_A = T_B = 80°F