Question
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60°C to 30 °C. The temprature of the surroundings is 20°C.
✓
Answer
According to Newton's law of cooling, we have
$\text{mc}\frac{(\text{T}_1-\text{T}_2)}{\text{t}}=\text{K}\big[\frac{(\text{T}_1+\text{T}_2)}{2}-\text{T}_0\big]\ ...(1)$
$\frac{\text{mc}}{\text{K}}=\frac{\big[\frac{\text{T}_1+\text{T}_2)}{2}-\text{T}_0\big]}{\big[\frac{(\text{T}_1-\text{T}_2)}{\text{t}}\big]}$
$\frac{\text{mc}}{\text{K}}=\frac{\big[\frac{(80+50)}{2}-20\big]}{\big[\frac{(80-50)}{5}\big]}=\frac{45}{6}$
Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in °C) in Eq. (1), we get
$\text{mc}\frac{(60-30)}{\text{t}}=\text{K}\Big[\frac{(60-30)}{2}-20\Big]$
$\frac{30\text{mc}}{\text{t}}=25\text{K}$
$\text{t}=\frac{30\text{mc}}{(25\text{K})}$
$=\big(\frac{30}{25}\big)\times\big(\frac{45}{6}\big)$
= 9min
The body takes 9 min to cool from 60°C to 30°C.
$\text{mc}\frac{(\text{T}_1-\text{T}_2)}{\text{t}}=\text{K}\big[\frac{(\text{T}_1+\text{T}_2)}{2}-\text{T}_0\big]\ ...(1)$
$\frac{\text{mc}}{\text{K}}=\frac{\big[\frac{\text{T}_1+\text{T}_2)}{2}-\text{T}_0\big]}{\big[\frac{(\text{T}_1-\text{T}_2)}{\text{t}}\big]}$
$\frac{\text{mc}}{\text{K}}=\frac{\big[\frac{(80+50)}{2}-20\big]}{\big[\frac{(80-50)}{5}\big]}=\frac{45}{6}$
Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in °C) in Eq. (1), we get
$\text{mc}\frac{(60-30)}{\text{t}}=\text{K}\Big[\frac{(60-30)}{2}-20\Big]$
$\frac{30\text{mc}}{\text{t}}=25\text{K}$
$\text{t}=\frac{30\text{mc}}{(25\text{K})}$
$=\big(\frac{30}{25}\big)\times\big(\frac{45}{6}\big)$
= 9min
The body takes 9 min to cool from 60°C to 30°C.
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