Answer to Question #128024 in Differential Equations for Giselle Ranchez

Question #128024

Suppose that two cups A and B are filled with coffee at the same time. Initially, the temperature of the coffee is 65° C. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of the coffee in cup A is 39° C. If Tm = 20° C,

 then what is the temperature of the coffee in cup B after 30 min?


1
Expert's answer
2020-08-04T19:25:38-0400

Solution.

"dT\/dt=ks(T-T_m)" ,  where k < 0

Here Tm= 20°C and S is exposed surface area.

Here Sa/Sb=2 , where SA and SB are the exposed surface area of A-cup and B-cup respectively.

Therefore

"dT\/((T-20))=ksdt"

Integrating,

"\u222bdT\/((T-20))=\u222bksdt"

=> ln(T-20) = kSt + C, C is integration constant.

Temperature are being measured in °C and time in minute here.

When t = 0 , initial temperature T = 65

So, ln(65 - 20) = C

=> C = ln(45)

So ln(T-20) = kSt + ln(45 )

So ln(T-20) - ln(45) = kSt

"ln (T-20)\/45=kSt"

For cup A, S = SA and T = 39 when t = 30

So equation for cup-A is

"ln (39-20)\/45=30kS_a"

"ln 19\/45=30kS_a" (1)

For cup-B, S = SB and T=? when t = 30

So equation for cup-B is

"ln (T-20)\/45=30kS_b"

"ln (T-20)\/45=60kS_a" (2) as SB =2 SA

Comparing eq(2) and eq(1) we get

"ln (T-20)\/45=2ln 19\/45"

"ln (T-20)\/45=ln\u2061(19\/45)^2"

"(T-20)\/45=\u2061(19\/45)^2"

T - 20 = 361/45

T = 20 + 8.02

T = 28.02

So the temperature of the coffee in cup B after 30 min is 28.02° C

Answer: the temperature of the coffee in cup B after 30 min is 28.02° C

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