A metal sphere cools from 80^ ^ C to 60^ ^ C in 10 minute. The time...

Using Newton's law of cooling in approximation form:

$\dfrac{T_1 - T_2}{t} =$ K \left( \dfrac{T_1 + T_2}{2} -T_0 \right)<br><br>For thefirst interval:

\dfrac{80 - 60}{10} =$K \left( \dfrac{80 + 60}{2} -30 \right)$

$2 = K (70 - 30)$

$2 = 40 K \Rightarrow K = \dfrac{1}{20}$

For the second interval:

$\dfrac{60 - 40}{t} =$ K \left( \dfrac{60 + 40}{2} -30\right)$<br><br>$\dfrac{20}{t} =$K (50 - 30)$

\dfrac{20}{t} =20 K$<br><br>Substituting the value of$K$:<br><br>$\dfrac{20}{t} =20 \times \dfrac{1}{20}

\dfrac{20}{t} =$1$

$t = 20$ minute

Answer: 20