Need help with these two problems Thanks A cup of hot coffee

Need help with these two problems. Thanks!

A cup of hot coffee was placed in a room maintained at a constant temperature of 73 degrees, and the coffee temperature was recorded periodically, in Table 1. Since y = C - 73, we have coffee temperature C - y + 73. Take your difference estimate from part (a) and add 73 degrees. Interpret the result by filling in the blank: When 35 minutes have elapsed, the estimated coffee temperature is degrees Suppose the coffee temperature C is 140 degrees. Then y = C - 73 = degrees is the temperature difference between the coffee and room temperatures. Consider the equation - 89.976 e^-0.023r where the is filled in with your answer from part (c). Show algebraic work to solve this part (d) equation for t, to the nearest tenth. Interpret your results clearly in the context of the coffee application. (Use additional paper if needed]

Solution

(a) Given, the temperature difference y = 89.976 e ^{-0.023 t}  

At t = 35 minutes, the temperature difference is

= 89.976 e ^-0.023x35 = 89.976 e ^-0.805 = 40.2 ^oC

(b) The estimated coffee temperature C = y + 73

= 40.2 + 73 = 113.2 ^oC

(c) Here, we have

C - 73 = 140 - 73 = 67 ⁰C

(d) From part (c), we have

      C - 73 = 89.976 e ^{-0.023 t}  (i.e. the value of y)

EXTRA CREDIT:

We have, C - 73 = 89.976 e ^{-0.023 t}

or,   e ^{-0.023 t} = (C - 73) / 89.976

Therefore, - 0.023 t = ln (C - 73) - ln 89.976

or, t = (ln 89.976) / 0.023 - (ln (C - 73)) / 0.023

t = 195.6 - 43.5 ln (C - 73)   [Values taken up to tength places of decimal]

The time t depend on the the coffee temperature. If the coffee temperature is high, then it will cool at a fast rate i.e. cooling will be faster.