The differential equation below models the temperature of an 88°C cup of coffee in a 24°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 74°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 88°C.) dy dt = − 1 50 (y − 24)

Explanation:

Let solve the differential equation by separating corresponding variables:

The solution of this equation is:

The explicit form of the temperature as a function of time is:

The value of the integration constant is:

The complete expression is:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j