Given, in first case, $T_1=150^{\circ} \mathrm{F}, T_2=144^{\circ} \mathrm{F}, t_1=1 \mathrm{~min}$ and room temperature, $T_0=72^{\circ} \mathrm{F}$
According to Newton's law of cooling,
$$
\begin{aligned}
& m c \frac{\left(T_1-T_2\right)}{t_1}=K\left(\frac{T_1+T_2}{2}-T_0\right) \\
& m c\left(\frac{150-144}{1}\right)=K\left(\frac{150+144}{2}-72\right) \\
& 6 m c=75 K
\end{aligned}
$$

In second case, $T^{\prime}{ }_1=110^{\circ} \mathrm{F}, T^{\prime}{ }_2=104^{\circ} \mathrm{F}$ Time $=t_2=$ ?
By Newton's law of cooling, we have
$$
\begin{aligned}
m c\left(\frac{110-104}{t_2}\right) & =K\left(\frac{110+104}{2}-72\right) \\
\frac{6 m c}{t_2} & =35 \mathrm{~K}
\end{aligned}
$$

From Eqs. (i) and (ii), we get,
$$
\begin{aligned}
& \frac{6 m c}{\left(\frac{6 m c}{t_2}\right)}=\frac{75 K}{35 K} \\
\Rightarrow \quad t_2=\frac{15}{7} & =2.14 \mathrm{~min} .
\end{aligned}
$$