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

The Correct Option is A

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 = e^C 
e^C = 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 = 60^e(-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.