Question

The temperature of the body drops from 60°C to 40°C in 7 min. The surrounding temperature is 10°C. The temperature of the body drops from 40°C to T°C in 7 min. Find the value of T

• A. 16°C
• B. 20°C
• C. 28°C
• D. 36°C

Answer 1

The Correct Option is C

The correct option is (C) 28°C
\(\frac{(60-40)}{7}=K(50-10)\)
\(\Rightarrow\frac{(40-T)}{7}=K([\frac{40+T}{2}]-10)\)
\(\Rightarrow\frac{20}{(40-T)}=\frac{(40\times2)}{(T+20)}\) 
\(\Rightarrow T+20=160-4T\)
\(\Rightarrow 5T = 140\)
\(T = \frac{140}{5}\) 
\(T\) = 28°C

Answer 2

Method-1:
Using the exact law of cooling: \[ T - T_s = (T_0 - T_s) e^{-kt} \] Case-I: \((40 - 10) = (60 - 10) e^{-7k}\) \[ 30 - 50e^{-7k} \quad \dots \, (1) \] Case-II: \((T - 10) = (40 - 10) e^{-7k} \quad T - 10 = 30e^{-7k}\) Dividing equation (2) by (1): \[ \frac{40 - T}{30} = \frac{20}{40 + T} \] \[ \Rightarrow T - 10 = 30 \times \frac{30}{50} = 18 \] Thus, the temperature after 7 more minutes is \(T = 28°C\). \bigskip Method-2:
Using Newton's law of cooling: \[ \frac{T - T_s}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right) \] Case-I: \[ \frac{60 - 40}{7} = k \left( \frac{40 + 10}{2} - 10 \right) \Rightarrow k = \frac{20}{7} \] Case-II: \[ \frac{40 - 10}{7} = k \left( \frac{20 + T}{2} - 10 \right) \Rightarrow T = 28°C \]