A body takes 10 minutes to cool from 60$$^\circ$$C to 50$$^\circ$$C. The temperature of surroundings is constant at 25$$^\circ$$C. Then, the temperature of the body after next 10 minutes will be approximately

We apply Newton’s law of cooling which states first: $$\frac{dT}{dt}=-k\,(T-T_s),$$ where $$T$$ is the temperature of the body at time $$t$$, $$T_s$$ is the constant temperature of the surroundings and $$k$$ is a positive constant characteristic of the system.

Integrating this differential equation, we obtain the standard formula

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

where $$T_0$$ is the initial temperature at $$t=0$$.

For the data given, the temperature of the surroundings is $$T_s=25^\circ\text{C}$$, and the body cools from $$T_0=60^\circ\text{C}$$ to $$T_1=50^\circ\text{C}$$ in the first $$10$$ minutes. We substitute these values into the formula to determine the constant $$k$$.

At $$t_1=10\ \text{min}$$, we have

$$T_1-T_s=(T_0-T_s)\,e^{-k t_1}.$$

Explicitly,

$$50-25=(60-25)\,e^{-10k}.$$

Simplifying the numbers gives

$$25=35\,e^{-10k}.$$

Dividing both sides by $$35$$,

$$\frac{25}{35}=e^{-10k}\quad\Longrightarrow\quad e^{-10k}=\frac{5}{7}.$$

Taking natural logarithm on both sides,

$$-10k=\ln\!\left(\frac{5}{7}\right)\quad\Longrightarrow\quad k=-\frac{1}{10}\,\ln\!\left(\frac{5}{7}\right)=\frac{1}{10}\,\ln\!\left(\frac{7}{5}\right).$$

Now we wish to find the temperature $$T_2$$ of the body after another $$10$$ minutes, i.e., at $$t_2=20\ \text{min}$$ from the start. Using the same cooling law,

$$T_2-T_s=(T_0-T_s)\,e^{-k t_2}.$$

Because $$t_2=20$$ and $$e^{-k t_2}=(e^{-10k})^2,$$ we write

$$T_2-T_s=(T_0-T_s)\,(e^{-10k})^2.$$

We already have $$e^{-10k}=\dfrac{5}{7}$$, so

$$(e^{-10k})^2=\left(\frac{5}{7}\right)^2=\frac{25}{49}.$$

Also, $$T_0-T_s=60-25=35.$$ Substituting these numbers,

$$T_2-25=35\;\frac{25}{49}.$$

Multiplying,

$$T_2-25=\frac{35\times25}{49}=\frac{35}{49}\times25=\frac{5}{7}\times25=\frac{125}{7}\approx17.857.$$

Adding the surroundings temperature back,

$$T_2=25+17.857\approx42.857^\circ\text{C}.$$

Rounding to the nearest whole number gives $$T_2\approx43^\circ\text{C}.$$

Hence, the correct answer is Option A.