Question

A pot of boiling water with a temperature of 100°C is set in a room with a temperature of 20°C. The temperature T of the water after x hours is given by T(x) = 20 + 80 e *. (a) Estimate the temperature of the water after 2 hours. (b) How long did it take the water to cool to 30°C? After 2 hours, the tempertaure of the water will be approximately (Type an integer or decimal rounded to one decimal place as needed.) The water will cool to 30°C in about hour(s). (Type an integer or decimal rounded to two decimal places as needed.)

Answer 1

After 2 hours, the water's temperature will be approximately 49.4°C, estimated using the provided decay function. It will take about 4.79 hours for the water to cool to 30°C, which is found by solving the temperature function for the desired temperature.

Step-by-step explanation:

To estimate the temperature of the water after 2 hours using the provided function T(x) = 20 + 80e^(-0.5x), we need to plug in the value of x as 2:

T(2) = 20 + 80e^(-0.5 × 2)

T(2) = 20 + 80e-1

Using a calculator, we find that e-1 is approximately 0.3679. Substituting this value in, we get:

T(2) = 20 + 80 × 0.3679

T(2) = 20 + 29.432

T(2) ≈ 49.4°C

To find how long it takes for the water to cool to 30°C, we need to solve the equation:

30 = 20 + 80e^(-0.5t)

Subtracting 20 from both sides gives us 10 = 80e^(-0.5t), or e^(-0.5t) = 0.125. Taking the natural logarithm of both sides, we find:

-0.5t = ln(0.125)

t = -2 × ln(0.125)

t ≈ 4.79

So, it will take approximately 4.79 hours for the water to cool to 30°C.

Answer 2

The temperature of the water after a certain time is determined using an exponential decay function. To estimate the temperature after 2 hours or find the time taken to cool to 30°C, we need the decay constant k, which is missing. Assuming k is provided, we plug it into the function to calculate the required values.

Step-by-step explanation:

When evaluating the temperature of water after a certain amount of time, we use the given exponential decay function T(x) = 20 + 80e^(*x), where T(x) is the temperature after x hours.

Part A: Estimating Temperature After 2 Hours

To estimate the temperature after 2 hours, we substitute x with 2 into the function:

T(2) = 20 + 80e^(-2k)

The value of "k" seems to be missing in the provided function. Assuming that there is a constant k involved, if we knew the value of k, we would plug it into the function to calculate the temperature. Without this value, we cannot accurately estimate the temperature after 2 hours.

Part B: Cooling Time to 30°C

To find out how long it takes for the water to cool to 30°C, we solve for x when T(x) = 30:

30 = 20 + 80e^(-kx)

Again, we would need the value of k to solve for x. Assuming k was provided, we would isolate e^(-kx) and then take the natural logarithm of both sides to solve for x.