Question

Suppose that a cup of soup cooled from 90 degrees C to 60 degrees C after 25 minutes in a room whose temperature was 20 degrees C. Use​ Newton's law of cooling to answer the following questions.
(A) How much longer would it take the soup to cool to 30 degree C?
(B) Instead of being left to stand in the room, the cup of 90 degree C soup is put in the freezer whose temperature is - 10 degree C. How long will it take the soup to cool from 90 degree C to 30 degree C?

Answer 1

Answer:

A)  61.93 minutes

B)  40.93 minutes

Step-by-step explanation:

Newton's law of cooling

$T(t)=T_S+(T_0-T_S)e^{-kt}$

where:

• T(t) is the temperature of the object at time t.
• $T_S$ is the temperature of the surrounding environment.
• $T_0$ is the initial temperature of the object.
• t is time.
• k is a constant.

Given values:

• $T_0$ = 90°C
• $T_S$ = 20°C
• t = 25 minutes
• T(t) = 60°C

Substitute the given values into the formula and solve for k:

$\implies 60=20+(90-20)e^{-25k}$

$\implies 40=70e^{-25k}$

$\implies e^{-25k}=\dfrac{4}{7}$

$\implies \ln e^{-25k}=\ln \dfrac{4}{7}$

$\implies -25k \ln e=\ln \dfrac{4}{7}$

$\implies -25k=\ln \dfrac{4}{7}$

$\implies k=-\dfrac{1}{25}\ln \dfrac{4}{7}$

Part A

Given values:

• $T_0$ = 60°C
• $T_S$ = 20°C
• $k=-\dfrac{1}{25}\ln \dfrac{4}{7}$
• T(t) = 30°C

Substitute the given values into the formula and solve for t:

$\implies 30=20+(60-20)e^{\frac{1}{25}t\ln \frac{4}{7}}$

$\implies 10=40e^{\frac{1}{25}t\ln \frac{4}{7}}$

$\implies e^{\frac{1}{25}t\ln \frac{4}{7}}=\dfrac{1}{4}$

$\implies \ln e^{{\frac{1}{25}t\ln \frac{4}{7}}}=\ln \dfrac{1}{4}$

$\implies {\dfrac{1}{25}t\ln \dfrac{4}{7}}\ln e=\ln \dfrac{1}{4}$

$\implies {\dfrac{1}{25}t\ln \dfrac{4}{7}}=\ln \dfrac{1}{4}$

$\implies t=\dfrac{\ln \dfrac{1}{4}}{{\frac{1}{25}\ln \frac{4}{7}}}$

$\implies t=61.93063129...$

$\implies t=61.93\;\; \sf minutes$

Part B

Given values:

• $T_0$ = 90°C
• $T_S$ = -10°C
• $k=-\dfrac{1}{25}\ln \dfrac{4}{7}$
• T(t) = 30°C

Substitute the given values into the formula and solve for t:

$\implies 30=-10+(90-(-10))e^{\frac{1}{25}t\ln \frac{4}{7}}$

$\implies 40=100e^{\frac{1}{25}t\ln \frac{4}{7}}$

$\implies e^{\frac{1}{25}t\ln \frac{4}{7}}=0.4$

$\implies \ln e^{\frac{1}{25}t\ln \frac{4}{7}}=\ln 0.4$

$\implies{\dfrac{1}{25}t\ln \dfrac{4}{7}}\ln e=\ln 0.4$

$\implies {\dfrac{1}{25}t\ln \dfrac{4}{7}}=\ln 0.4$

$\implies t=\dfrac{\ln 0.4}{{\frac{1}{25}\ln \frac{4}{7}}}$

$\implies t=40.93392072...$

$\implies t=40.93\;\; \sf minutes$