Answer:
a). 59.049°C
b). 2.1179 seconds
Step-by-step explanation:
Expression representing the final temperature after decrease in temperature of the metal from 100°C to T°C is,
T = [tex]100(0.9)^{x}[/tex]
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 = [tex]100(0.9)^{x}[/tex]
0.8 = [tex](0.9)^{x}[/tex]
By taking log on both the sides
log(0.8) =log[[tex](0.9)^{x}[/tex]]
-0.09691 = x[log(0.9)]
-0.09691 = -0.045757x
x = [tex]\frac{0.09691}{0.045757}[/tex]
x = 2.1179
x ≈ 2.1179 seconds
