The Temperature of a Cup of Coffee Obeys Newton's Law $140 ^ { \circ } \mathrm { F }$

The temperature of a cup of coffee obeys Newton's law of cooling. The initial temperature of the coffee is $140 ^ { \circ } \mathrm { F }$ and one minute later, it is $125 ^ { \circ } \mathrm { F }$ . The ambient temperature of the room is $65 ^ { \circ } \mathrm { F }$ . If $T ( t )$ represents the temperature of the coffee at time t, the correct differential equation for the temperature is

A) $\frac { d T } { d t } = k ( T - 125 )$
B) $\frac { d T } { d t } = k ( T - 140 )$
C) $\frac { d T } { d t } = k ( T - 65 )$
D) $\frac { d T } { d t } = T ( T - 140 )$
E) $\frac { d T } { d t } = T ( T - 65 )$