Question

a body originally at 80°C cools to 60°C on 20mins the temperature of the air being 40°C what will be the temperature of the body after 40 mins​

Answer 1

Answer:

The temperature of the body after 40 mins​ is 44°C.

Explanation:

From the Newtons law of cooling we have,

$\frac{T_f-T_i}{\Delta t}=K[\frac{T_f+T_i}{2}-T_a]$         (1)

From the question we have,

$T_f$=80°C

$T_i$=60°C

Δt₁=20min

Δt₂=40min

By substituting the values in equation (1) we get;

$\frac{80\textdegree-60\textdegree}{20}=K[\frac{80\textdegree+60\textdegree}{2}-40\textdegree]$

$K=\frac{1}{30}$         (2)

Now as per the question equation (1) becomes,

$\frac{60\textdegree-T_i}{40}=K[\frac{60\textdegree+T_i}{2}-40\textdegree]$     (3)

By using equation (2) in equation (3) we get;

$\frac{60\textdegree-T_i}{40}=\frac{1}{30} [\frac{60\textdegree+T_i}{2}-40\textdegree]$

$180\textdegree-3T_i=120\textdegree+2T_i-160\textdegree$

$5T_i=(180+160-120)\textdegree C$

$T_i=\frac{220\textdegree C}{5}=44\textdegree C$

Hence, the temperature of the body after 40 mins​ is 44°C.