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The Temperature of a Cup of Coffee Obeys Newton's Law 150F150 ^ { \circ } \mathrm { F }

Question 23

Multiple Choice

The temperature of a cup of coffee obeys Newton's law of cooling. The initial temperature of the coffee is 150F150 ^ { \circ } \mathrm { F } and one minute later, it is 135F135 ^ { \circ } \mathrm { F } . The ambient temperature of the room is 70F70 ^ { \circ } \mathrm { F } . If T(t) T ( t ) represents the temperature of the coffee at time t, the correct differential equation for the temperature with side conditions is


A) dTdt=k(T135) \frac { d T } { d t } = k ( T - 135 )
B) dTdt=k(T150) \frac { d T } { d t } = k ( T - 150 )
C) dTdt=k(T70) \frac { d T } { d t } = k ( T - 70 )
D) dTdt=T(T150) \frac { d T } { d t } = T ( T - 150 )
E) dTdt=T(T70) \frac { d T } { d t } = T ( T - 70 )

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