Question

The temperature of a cup of a coffee is 92°c, when freshly pour the room temperature 24°c . In 1min it was cool to 80°c how long a period must alapse before the temperature of cup becomes 65°c​

Answer 1

Answer:

when the room temperature is 30 degree Celsius then the coffee temperature will be 65 degree Celsius

Answer 2

Answer: This is a Newton's Law of Cooling application. The coffee is the substance being cooled and the atmosphere is the cooling medium.

Step-by-step explanation:STEP:1 Newton's Law of Cooling Reference

Here is the Newton’s Law of Cooling equation for temperature  T(t)  over time. This is the solution to a differential equation but Calculus isn’t needed to solve this.

T(t)=TS+(T0−TS)e−kt

T0=200∘  is the initial coffee temperature.

TS=70∘  is the surrounding air temperature.

k=?  is a constant given by the experiment.

Compute constant  k . The question gives experimental results over one minute’s time.

The coffee has a temperature of 200°F when freshly poured and 1 minute later has cooled to 190°F in a room at 70°F. Enter these values into the equation and solve for  k .

T(t)=TS+(T0−TS)e−kt

T(1)=190=70+(200−70)ek×(−1)

Recall that  (−1)ln(x)=ln(1x)  for the next line.

120=130e−k⟺k=(−1)ln(1213)⟺k=ln(1312)

Compute the time  t  when the coffee reaches the temperature 150°F.

T(t)=TS+(T0−TS)e−kt

T(t)=70+130e−tln(1312)  is the general function.

Set the temperature to 150°F and solve for  t .

150=70+130e−tln(1312)

t=ln(138)ln(1312)≈6.067  minutes

STEP:2

This is an experiment I repeat daily, though never quite to the end, as I like drinking my hot beverage while still hot.

It is, however, amazing how long hot water, which is really what we are talking about here, stays hot. When I make myself a nice 1.5 litre pot of Assam tea (I repeat, the cooling dynamics will be the same as coffee), the following times go by:

1. letting it steep, 20 minutes;

2. taking about 40 minutes to drink it, and the last cup will still be perfectly accepted..

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