Question

Через сколько времени температура тела нагретого до 100 , понизится до 25 C, если температура помещения равна 20 C и за первые 10 мин тело охладилось до 60 C?

Answer 1

Если это не физика, а линейная математическая прогрессия, то
1) 100 - 25 = 75 град - нужно охладить
2) 100 - 60 = 40 град - охладилось за 10 минут
3)  Пропорция    10 мин : 40 град = Х мин : 75 град
Отсюда
Х = 10 * (75/40) = 18,75 мин = 18 мин 45 сек - ОТВЕТ

Answer 2

Calculation of Body Temperature Decrease

To calculate the time it takes for the body temperature to decrease from 100°C to 25°C, we need to consider the initial temperature of the body, the temperature of the room, and the rate at which the body cools down.

Given: - Initial body temperature: 100°C - Room temperature: 20°C - Body temperature after 10 minutes: 60°C

We can assume that the body temperature decreases exponentially over time. The rate of cooling can be described by Newton's law of cooling, which states that the rate of heat loss is proportional to the temperature difference between the body and its surroundings.

Using this information, we can set up the following equation:

T(t) = T_room + (T_initial - T_room) * e^(-kt)

Where: - T(t) is the body temperature at time t - T_room is the room temperature - T_initial is the initial body temperature - k is the cooling constant - e is the base of the natural logarithm

To solve for the cooling constant k, we can use the given information that the body temperature after 10 minutes is 60°C:

60 = 20 + (100 - 20) * e^(-10k)

Simplifying the equation, we get:

e^(-10k) = 40/80 = 1/2

Taking the natural logarithm of both sides, we have:

-10k = ln(1/2)

Solving for k, we get:

k = -ln(1/2)/10 ≈ 0.0693

Now that we have the cooling constant, we can find the time it takes for the body temperature to decrease from 100°C to 25°C. We can set up the equation as follows:

25 = 20 + (100 - 20) * e^(-0.0693t)

Simplifying the equation, we get:

5 = 80 * e^(-0.0693t)

Dividing both sides by 80, we have:

1/16 = e^(-0.0693t)

Taking the natural logarithm of both sides, we get:

ln(1/16) = -0.0693t

Solving for t, we have:

t = ln(1/16)/(-0.0693) ≈ 19.9 minutes

Therefore, it takes approximately 19.9 minutes for the body temperature to decrease from 100°C to 25°C, given the initial conditions and the rate of cooling.