The initial temperature of a body is 80^ ^ C . If its temperature...

According to the exact form of Newton's law of cooling, the temperature $T$ at time $t$ is given by:

$T - T_0 = (T_i - T_0)e^{-Kt}$

where $T_0$ is the surrounding temperature and $T_i$ is the initial temperature.

For the first interval of 5 minutes:
T_i = 80^{}^{\circ}\text{C}, T = 64^{}^{\circ}\text{C}, $t = 5$
$64 - T_0 = (80 - T_0)e^{-5K}$
$e^{-5K} = \dfrac{64 - T_0}{80 - T_0}$ --- (1)

For the next interval of 10 minutes:
T_i = 64^{}^{\circ}\text{C}, T = 52^{}^{\circ}\text{C}, $t = 10$
$52 - T_0 = (64 - T_0)e^{-10K}$
$e^{-10K} = \dfrac{52 - T_0}{64 - T_0}$ --- (2)

Since $e^{-10K} = (e^{-5K})^2$, equating (2) and the square of (1):
$\dfrac{52 - T_0}{64 - T_0} = \left(\dfrac{64 - T_0}{80 - T_0}\right)^2$

To simplify, let $y = 64 - T_0$. Then $52 - T_0 = y - 12$ and $80 - T_0 = y + 16$.

$\dfrac{y - 12}{y} = \left(\dfrac{y}{y + 16}\right)^2$
$(y - 12)(y + 16)^2 = y^3$
$(y - 12)(y^2 + 32y + 256) = y^3$
$y^3 + 32y^2 + 256y - 12y^2 - 384y - 3072 = y^3$
$20y^2 - 128y - 3072 = 0$
$5y^2 - 32y - 768 = 0$

Solving for $y$ using the quadratic formula:
$y = \dfrac{32 \pm \sqrt{(-32)^2 - 4(5)(-768)}}{2(5)}$
$y = \dfrac{32 \pm \sqrt{1024 + 15360}}{10}$
$y = \dfrac{32 \pm \sqrt{16384}}{10}$
$y = \dfrac{32 \pm 128}{10}$

Taking the positive root (since T_0 < 52^{}^{\circ}\text{C}, $y > 12$):
$y = \dfrac{160}{10} = 16$

Substituting back $y = 64 - T_0$:
$16 = 64 - T_0$
T_0 = 48^{}^{\circ}\text{C}

Answer: 48^{}^{\circ}\text{C}