According to Newton, the rate at which the temperature of water in a swimming pool changes is directly proportional to the difference between the temperature L outside and the temperature T of the water in the pool. Suppose the temperature outside stays at a constant 28 C for two hours, and that during that time the temperature of the water increases from 20 C to 24 C. Which of the following shows the differential equation for this situation and its solution?
A) ;
B) ;
C) ;
D) ;

Question

According to Newton, the rate at which the temperature of water in a swimming pool changes is directly proportional to the difference between the temperature L outside and the temperature T of the water in the pool. Suppose the temperature outside stays at a constant 28  C for two hours, and that during that time the temperature of the water increases from 20  C to 24  C. Which of the following shows the differential equation for this situation and its solution?
A) ; 
B) ; 
C) ; 
D) ;

Quizplus · Accepted Answer

Let T(t) be the temperature of the water at time t (in hours, since Newton specified the rate of change was in units of temperature per hour). Then the rate of change of temperature is dT/dt. By Newton's law, we have:

dT/dt = k(L - T)

where k is the constant of proportionality.

We are given L = 28 11eb69ff_fde5_1db6_af94_61cd33f6f89d_TB8190_11 C and initial condition T(0) = 20 11eb69ff_fde5_1db7_af94_b3a7f1b81cd3_TB8190_11 C.

To solve the differential equation, we can separate variables and integrate:

dT/(L-T) = k dt

Integrating both sides, we get:

-ln(L-T) = kt + C

where C is a constant of integration. Exponentiating both sides, we get:

L - T = Ce^(-kt)

Rearranging, we get:

T(t) = L - Ce^(-kt)

Using the initial condition T(0) = 20 11eb69ff_fde5_1db7_af94_b3a7f1b81cd3_TB8190_11 C, we can solve for C:

20 11eb69ff_fde5_1db7_af94_b3a7f1b81cd3_TB8190_11 = 28 11eb69ff_fde5_1db6_af94_61cd33f6f89d_TB8190_11 - Ce^(0)

C = 8

So the solution is:

T(t) = 28 11eb69ff_fde5_1db6_af94_61cd33f6f89d_TB8190_11 - 8e^(-kt)

To find the value of k, we use the fact that the temperature increases from 20 11eb69ff_fde5_1db7_af94_b3a7f1b81cd3_TB8190_11 C to 24 11eb69ff_fde5_1db8_af94_67b41c4f05a8_TB8190_11 C over the 2-hour period. That is:

T(2) = 24 11eb69ff_fde5_1db8_af94_67b41c4f05a8_TB8190_11 C

Substituting into our solution, we have:

24 11eb69ff_fde5_1db8_af94_67b41c4f05a8_TB8190_11 C = 28 11eb69ff_fde5_1db6_af94_61cd33f6f89d_TB8190_11 - 8e^(-2k)

Solving for k, we get k ≈ 0.693/h (to three decimal places).

Therefore, the differential equation and its solution are: 
dT/dt = 0.693(L-T)
T(t) = 28 11eb69ff_fde5_1db6_af94_61cd33f6f89d_TB8190_11 - 8e^(-0.693t)

According to Newton, the Rate at Which the Temperature of Water