A hot brick is removed from a kiln at 250C above room temper

A hot brick is removed from a kiln at 250C above room temperature. Over time, the brick cools off. After 2 hours have elapsed, the brick is 10C above room temperature. Let t be the time in hours since the brick was removed from the kiln. Let y=H(t) be the difference between the brick\'s and the room\'s temperature at time t. Assume that H(t) is an exponential function. (1 point) A hot brick is removed from a kiln at 250C above room temperature. Over time, the brick cools off. After 2 hours have elapsed, the brick is 10C above room temperature. Let t be the time in hours since the brick was removed from the kiln. Let y=H(t) be the difference between the brick\'s and the room\'s temperature at time t. Assume that H(t) is an exponential function. (a) Find a formula for H(t). H(t)= . (b) How many degrees does the brick\'s temperature drop during the first quarter hour? It drops (c) How many degrees does the temperature drop during the next quarter hour? It drops (d) Find H1(y). H1(y)= (e) How much time elapses before the brick\'s temperature is 3C above room temperature? time = ) hours.

Solution

formula would be H(t) = 250 e^ (kt)

after 2 hrs H(t) = 10

So, 10 = 250e^(k*2)

(1/25) = e^(2k)

taking natural log on both sides:

ln(1/25) = 2k -----> k = -1.61

a) So, H(t) = 250e^(-1.6t)

b) brick\'s temperature drop during the first quarter hour

t = 0.25 hr

So, H(t) e^(-1.6*0.25) = 250e^( -0.4) = 250*0.67

= 167.5 deg

c) in next quarter hour first find how muc drop was in first half hour

H(t) = 250e^(-1.6*0.5) = 118.1 deg

Now to find drop during the next quarter hour = 118.1 - 167.5 = -49.4

drop during the next quarter hour = 49.4 deg

d)   H(t) = 250e^(-1.6t)

find inverse of H(t)

plug t= H(t) and H(tr=) = t in the above equation and solve for H

t = 250e^-1.6H

take log on both sides:

ln(t) = ln250 -1.6H

1.6H = ln(250) - lnt

H = (1/1.6)ln(250/t)