Question

Please help this question.

The temperature of a type of metal decreases from 100°C to T°C according to T = 100(0.9)*
after x second. Calculate
(a) the temperature of the metal after 5 seconds,
(b) the time taken, x, in seconds for the temperature of the metal to decrease from 100°C to
80°C.​

Answer 1

a). 59.049°C

b). 2.1179 seconds

Explanation:

Expression representing the final temperature after decrease in temperature of the metal from 100°C to T°C is,

T =
`100(0.9)^(x)`

where x = duration of cooling

a). Temperature when x = 5 seconds

T = 100(0.9)⁵

= 59.049

≈ 59.049°C

b). If the temperature of the metal decreases from 100°C to 80°C

Which means for T = 80°C we have to calculate the duration of cooling 'x' seconds

80 =
`100(0.9)^(x)`

0.8 =
`(0.9)^(x)`

By taking log on both the sides

log(0.8) =log[
`(0.9)^(x)`]

-0.09691 = x[log(0.9)]

-0.09691 = -0.045757x

x =
`(0.09691)/(0.045757)`

x = 2.1179

x ≈ 2.1179 seconds