A body cools from $$60°$$C to $$40°$$C in $$6$$ minutes. If, temperature of surroundings is $$10°$$C. Then, after the next 6 minutes, its temperature will be ______ °C.

A body cools from 60°C to 40°C in 6 minutes. Temperature of surroundings is 10°C. Find the temperature after the next 6 minutes.

According to Newton's Law of Cooling: $$T(t) - T_s = (T_0 - T_s)e^{-kt}$$ where $$T_s = 10°$$C is the surrounding temperature.

At $$t=6$$ min, $$T=40°$$C, $$T_0=60°$$C, so $$40-10=(60-10)e^{-6k}$$, giving $$30=50e^{-6k}$$ and hence $$e^{-6k}=\frac{3}{5}$$.

At $$t=12$$ min from the start, the temperature satisfies $$T-10=50e^{-12k}=50(e^{-6k})^2=50\left(\frac{3}{5}\right)^2=18$$, so $$T=28°\text{C}$$.

Verification using the interval method: Alternatively, treating the second interval separately with initial temperature 40°C: $$T-10=(40-10)e^{-6k}=30\cdot\frac{3}{5}=18$$, hence $$T=28°\text{C}$$ ✓

Answer: 28°C