Question

1. A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law

Answer 1

Answer:

a) The inital temperature of the soup is 183 ºF.

b) The temperature of the soup after 10 minutes is approximately 68.261 ºF.

c) The temperature of the soup will be 100 ºF in approximately 4.363 minutes.

d) We kindly invite you to see the image attached below.

Step-by-step explanation:

Let the temperature of the bowl of soul be represented by this solution according to Newton's Law of Cooling:

$T(t) = 58+125\cdot e^{-0.25\cdot t}$ (1)

Where:

$t$ - Time, measured in minutes.

$T(t)$ - Temperature, measured in degrees Fahrenheit.

a) We find the initial temperature of the bowl of soup by evaluating (1) at $t = 0\,min$:

$T(0) = 58+125$

$T(0) = 183\,^{\circ}F$

The inital temperature of the soup is 183 ºF.

b) We find the temperature of the bowl of soup by evaluating (1) at $t = 10\,min$:

$T(10) = 58+125\cdot e^{-0.25\cdot (10)}$

$T(10) \approx 68.261\,^{\circ}F$

The temperature of the soup after 10 minutes is approximately 68.261 ºF.

c) If we know that $T(t) = 100\,^{\circ}F$, then the instant associated with such temperature is:

$100 = 58+125\cdot e^{-0.25\cdot t}$

$125\cdot e^{-0.25\cdot t} = 42$

$e^{-0.25\cdot t} = \frac{42}{125}$

$-0.25\cdot t = \ln \frac{42}{125}$

$t = -\frac{1}{0.25}\cdot \ln \frac{42}{125}$

$t \approx 4.363\,min$

The temperature of the soup will be 100 ºF in approximately 4.363 minutes.

d) At last we sketch the graph of the temperature versus time.