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A cup of coffee cools from 90 ∘ C 90^{\circ}C 9 0 ∘ C to 80 ∘ C 80^{\circ}C 8 0 ∘ C in t t t minutes, when the room temperature is 20 ∘ C . 20^{\circ}C. 2 0 ∘ C . The time taken by a similar cup of coffee to cool from 80 ∘ C 80^{\circ}C 8 0 ∘ C to 60 ∘ C 60^{\circ}C 6 0 ∘ C at a room temperature same at 20 ∘ C 20^{\circ}C 2 0 ∘ C is:
hard
Thermal Properties of Matter
2021
physics
13 10 t \frac{13}{10} t 10 13 t
Explanation To solve this problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature. \\
The formula for Newton's Law of Cooling is: d T d t = − k ( T − T r o o m ) \\
\frac{dT}{dt} = -k(T - T_{room}) \\
d t d T = − k ( T − T roo m )
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10 13 t \frac{10}{13} t 13 10 t
Where:
•
is the temperature of the object at time
•
is the ambient (room) temperature.
•
is a positive constant.
Given:
• Initial temperature
T 1 = 90 ∘ C , T_1 = 90^{\circ}C, T 1 = 9 0 ∘ C , final temperature
T 2 = 80 ∘ C . T_2 = 80^{\circ}C. \\
T 2 = 8 0 ∘ C . • Room temperature
T r o o m = 20 ∘ C . T_{room} = 20^{\circ}C. \\
T roo m = 2 0 ∘ C . • Time taken to cool from
to
is
minutes.
Using the formula:
T 2 − T r o o m T 1 − T r o o m = e − k t \\
\frac{T_2 - T_{room}}{T_1 - T_{room}} = e^{-kt} \\
T 1 − T roo m T 2 − T roo m = e − k t Substitute the known values:
80 − 20 90 − 20 = e − k t 60 70 = e − k t 6 7 = e − k t \\
\frac{80 - 20}{90 - 20} = e^{-kt} \\
\frac{60}{70} = e^{-kt} \\
\frac{6}{7} = e^{-kt} \\
90 − 20 80 − 20 = e − k t 70 60 = e − k t 7 6 = e − k t Now, for the second scenario:
• Initial temperature
T 3 = 80 ∘ C , T_3 = 80^{\circ}C, T 3 = 8 0 ∘ C , final temperature
T 4 = 60 ∘ C . T_4 = 60^{\circ}C. \\
T 4 = 6 0 ∘ C . We need to find the time taken, say
for this cooling process.
Using the same formula:
T 4 − T r o o m T 3 − T r o o m = e − k t 2 \\
\frac{T_4 - T_{room}}{T_3 - T_{room}} = e^{-kt_2} \\
T 3 − T roo m T 4 − T roo m = e − k t 2 Substitute the known values:
60 − 20 80 − 20 = e − k t 2 40 60 = e − k t 2 2 3 = e − k t 2 \\
\frac{60 - 20}{80 - 20} = e^{-kt_2} \\
\frac{40}{60} = e^{-kt_2} \\
\frac{2}{3} = e^{-kt_2} \\
80 − 20 60 − 20 = e − k t 2 60 40 = e − k t 2 3 2 = e − k t 2 Now, we have two equations:
e − k t = 6 7 e − k t 2 = 2 3 \\
e^{-kt} = \frac{6}{7} \\
e^{-kt_2} = \frac{2}{3} \\
e − k t = 7 6 e − k t 2 = 3 2 Taking the natural logarithm of both sides:
− k t = ln ( 6 7 ) − k t 2 = ln ( 2 3 ) \\
-kt = \ln\left(\frac{6}{7}\right) \\
-kt_2 = \ln\left(\frac{2}{3}\right) \\
− k t = ln ( 7 6 ) − k t 2 = ln ( 3 2 ) Divide the second equation by the first:
− k t 2 − k t = ln ( 2 3 ) ln ( 6 7 ) t 2 = t ⋅ ln ( 2 3 ) ln ( 6 7 ) \\
\frac{-kt_2}{-kt} = \frac{\ln\left(\frac{2}{3}\right)}{\ln\left(\frac{6}{7}\right)} \\
t_2 = t \cdot \frac{\ln\left(\frac{2}{3}\right)}{\ln\left(\frac{6}{7}\right)} \\
− k t − k t 2 = l n ( 7 6 ) l n ( 3 2 ) t 2 = t ⋅ l n ( 7 6 ) l n ( 3 2 ) Using logarithm properties, we approximate:
ln ( 2 3 ) ≈ − 0.4055 ln ( 6 7 ) ≈ − 0.1542 t 2 = t ⋅ − 0.4055 − 0.1542 t 2 ≈ t ⋅ 13 5 \\
\ln\left(\frac{2}{3}\right) \approx -0.4055 \\
\ln\left(\frac{6}{7}\right) \approx -0.1542 \\
t_2 = t \cdot \frac{-0.4055}{-0.1542} \\
t_2 \approx t \cdot \frac{13}{5} \\
ln ( 3 2 ) ≈ − 0.4055 ln ( 7 6 ) ≈ − 0.1542 t 2 = t ⋅ − 0.1542 − 0.4055 t 2 ≈ t ⋅ 5 13 Therefore, the time taken by a similar cup of coffee to cool from
to
is
13 5 t . \frac{13}{5} t. \\
5 13 t . This corresponds to Option 2.