Question

If surrounding air is kept at 20 °C and body cools from 80 °C to 70 °C in 5 minutes, then the temperature of the body after 15 minutes will be

• A. 54.7 °C
• B. 51.7 °C
• C. 52.7 °C
• D. 50.7 °C

Answer 1

Answer: (A)

54.7 °C

Answer 2

Let's denote the temperature of the body at time t as T(t) and the temperature of the surrounding air as T0. We can write the differential equation for the cooling process as:
$\frac {dT}{dt}$ = -k(T - T0)
Where k is the cooling constant.
Given that the body cools from 80 °C to 70 °C in 5 minutes, we can use this information to find the value of k. Let's use the midpoint temperature (75 °C) during this 5-minute interval:
(-k)(75 - 20) = 70 - 20
-55k = -50
k = $\frac {50}{55}$
K = $\frac {10}{11}$
Now we can solve the differential equation to find the temperature of the body after 15 minutes:
$\frac {dT}{dt}$ = -$\frac {10}{11}$ (T - 20)
Separating variables and integrating:
$\frac {1}{T-20}$ dT = -$\frac {10}{11}$ dt
Integrating both sides:
ln|T - 20| = $-\frac {10}{11}$t + C
Taking the exponential of both sides:
|T - 20| = e((-10/11)t + C)
Since T - 20 cannot be negative, we can remove the absolute value sign:
T - 20 = e((-10/11)t + C)
Simplifying the constant of integration, let's assume T(0) = 80 °C:
80 - 20 = eC
eC = 60
Substituting back into the equation:
T - 20 = 60e((-10/11)t)
Now we can find the temperature after 15 minutes (t = 15):
T - 20 = 60e((-10/11) * 15)
T - 20 = 60e(-150/11)
T = 20 + 60e(-150/11)
Calculating this expression, we find that T ≈ 54.7 °C.
Therefore, the temperature of the body after 15 minutes will be approximately 54.7 °C. Hence, the correct answer is option (A) 54.7 °C.