Question

A cup of coffee is poured from a pot, whose contents are ${95}^{o}C$ into a non-insulated cup in a room at ${20}^{o}C$. After a minute, the coffee has cooled to ${90}^{o}C$. How much time is required before the coffee reaches a drinkable temperature of ${65}^{o}C$ ?

• A. $4.3$ minutes
• B. $5.7$ minutes
• C. $6.0$ minutes
• D. $7.4$ minutes

Answer 1

The correct option is D $7.4$ minutes

Let $T$ be the temperature of the soup and $T_0$ be the temperature of the surroundings.

The Difference $D=T-T_0$. In this case $T_0=20$

By Newton's law, $dD=kDdt$

$\Rightarrow \int \frac{dD}{D}=\int kdt$

$\Rightarrow lnD=kt+c$ $....\left(1\right)$

At $t=0,T=95$ so $D=95-20=75$

$\Rightarrow ln\left(75\right)=k\left(0\right)+c$

$\therefore c=ln\left(75\right)$

Substitute $c$ in $\left(1\right)$ we get

$\Rightarrow lnD=kt+ln\left(75\right)$

$\Rightarrow lnD-ln\left(75\right)=kt$

$\Rightarrow ln\frac{D}{75}=kt$

At $t=1$ minute, $T=90$ so $D=70$

$\Rightarrow ln\frac{70}{75}=k\left(1\right)$

$\therefore k=-0.06899$

$\Rightarrow ln\frac{D}{75}=-0.06899t$ $....\left(2\right)$

$\Rightarrow \frac{D}{75}=e^{-0.06899t}$

$\Rightarrow D=75e^{-0.06899t}$

Coffee is drinkable when $T={65}^{\circ }C$ so $D=45$

Substitute in $\left(2\right)$ we get

$\Rightarrow ln\frac{45}{75}=-0.06899t$

$\Rightarrow t=7.4$ minutes.