Question

A cold drink is taken from a cooler. Initially, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C.

(a) What is the temperature of the drink after 45 minutes? (Round your answer to two decimal places.)
12.77 °C (CORRECT)

(b) When will its temperature be 14°C? (Round your answer to two decimal places.)
56.56 minutes (INCORRECT)

Answer 1

(a) the temperature of the drink after 45 minutes is 12.77
`^(o)C`

(b) its temperature be 14
`^(o)C` at t = 56.50 Minutes

Explanation:

Given information:

Initial temperature,
`T_(0)` = 5
`^(o)C`

Sorrounding temperature, C = 20
`^(o)C`

in 25 minute, T(25) = 10
`^(o)C`

(a) the temperature of the drink after 45 minutes

According Newton's law of cooling

T(t) = C + (
`T_(0)` - C)
`e^(-kt)`

where

`T_(0)` = initial temperature

t = time

k = cooling constant

C = sorrounding temperature

Thus,

T(25) = 20 + (5 - 20)
`e^(-25k)`

10 = 20 - (15
`e^(-25k)`)

`e^(-25k)` =
`(20-10)/(15)`

-25k = ln(
`(20-10)/(15)`)

k = - (ln(
`(20-10)/(15)`))/25

= 0.016

hence, the complete equation

T(t) = 20 + (5 - 20)
`e^(-0.016t)`

in 45 minute the temperature is

T(45) = 20 + (5 - 20)
`e^(-0.016 x 45)`

T(45) = 12.77
`^(o)C`

the temperature is 14
`^(o)C`, so T(t) = 14
`^(o)C`

T(t) = 20 + (5 - 20)
`e^(-0.016t)`

14 = 20 + (5 - 20)
`e^(-0.016t)`

`e^(-0.016t)` =
`(20-14)/(15)`

-0.016t = ln (
`(20-14)/(15)`)

t = - (ln (
`(20-14)/(15)`))/ 0.016)

t = 56.50 Minutes