Question

The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modification of (2), given in Section 3.1, is dT/dt = kS(T-Tm), where
k < 0
and Tm is a constant. 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 38° C. If Tm = 21° C, then what is the temperature of the coffee in cup B after 30 min? (Round your answer to two decimal places.)

______ degrees Celsius

Answer 1

$\frac {dT}{dt}$ = kS(T-T_m) where k < 0

Here T_m= 21°C and S is exposed surface area.

Here $\frac {S_B}{S_A} = 2$ , where S_A and S_B are the exposed surface area of A-cup and B-cup respectively.

Therefore

$\frac {dT}{(T - 21)}$ = kSdt

Integrating,

$\int$ $\frac {dT}{(T - 21)}$ = $\int$ kS dt

=> ln(T-21) = 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 - 21) = C

=> C = ln(44)

So ln(T-21) = kSt + ln(44 )

So ln(T-21) - ln(44) = kSt

=> ln $\frac {T-21}{44}$ = kSt

For cup A, S = S_A and T = 38 when t = 30

So equation for cup-A is

ln $\frac {38-21}{44}$ = 30kS_A

=> ln $\frac {17}{44}$ = 30kS_A •••••••••eq(1)

For cup-B, S = S_B and T=? when t = 30

So equation for cup-B is

ln $\frac {T-21}{44}$ = 30kS_B

ln $\frac {T-21}{44}$ = 60kS_A as S_B =2 S_A ••••eq(2)

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

ln $\frac {T-21}{44}$ = 2 ln $\frac {17}{44}$

=> ln $\frac {T-21}{44}$ = ln ($\frac {17}{44}$ )²

=> $\frac {T-21}{44}$ = $(\frac {17}{44})^2$

=> T - 21 = $\frac {289}{44}$

=> T = 21 + 6.568

=> T = 27.568

The temperature of coffee in cup B is 27.57 °C ( correct up to two decimal places)