Question

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 T_m = 20° C,

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

Answer 1

Solution.

$dT/dt=ks(T-T_m)$ ,  where k < 0

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

Here S_a/S_b=2 , where S_A and S_B are the exposed surface area of A-cup and B-cup respectively.

Therefore

$dT/((T-20))=ksdt$

Integrating,

$∫dT/((T-20))=∫ksdt$

=> 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 = S_A 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 = S_B 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 S_B =2 S_A

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

$ln (T-20)/45=2ln 19/45$

$ln (T-20)/45=ln⁡(19/45)^2$

$(T-20)/45=⁡(19/45)^2$

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