Question

show some problems like a body goes from 100C to 80C in time t when sorrounding temp is 25 then time required fro the body to go from 80 to 70

Answer 1

$\Delta t = \frac{t\,\ln(11/9)}{\ln(15/11)}$

Answer 2

We use Newton’s law of cooling, which states:

$$\frac{dT}{dt}=-k\,(T-T_s)$$

with solution

$$T - T_s = (T_0-T_s)e^{-kt}.$$

Step 1. Cooling from 100°C to 80°C

Let $T_0 = 100^\circ\text{C}$, $T = 80^\circ\text{C}$, and $T_s = 25^\circ\text{C}$. Then,

$$80 - 25 = (100-25)e^{-kt}\quad\Longrightarrow\quad 55 = 75\,e^{-kt}.$$

Thus,

$$e^{-kt}=\frac{55}{75}=\frac{11}{15}.$$

Taking logarithms:

$$kt = -\ln\left(\frac{11}{15}\right)=\ln\left(\frac{15}{11}\right).$$

So,

$$k=\frac{\ln(15/11)}{t}.$$

Step 2. Cooling from 80°C to 70°C

Now, for cooling from $T_0 = 80^\circ\text{C}$ to $T=70^\circ\text{C}$ with the same ambient $T_s = 25^\circ\text{C}$:

$$70 - 25 = (80-25)e^{-k\Delta t}\quad\Longrightarrow\quad 45 = 55\,e^{-k\Delta t}.$$

Thus,

$$e^{-k\Delta t}=\frac{45}{55}=\frac{9}{11}.$$

Taking logarithms:

$$k\Delta t = -\ln\left(\frac{9}{11}\right)=\ln\left(\frac{11}{9}\right).$$

Substitute $k=\frac{\ln(15/11)}{t}$:

$$\frac{\ln(15/11)}{t}\Delta t = \ln\left(\frac{11}{9}\right),$$

which gives

$$\Delta t = \frac{t\,\ln(11/9)}{\ln(15/11)}.$$