Question 686: Thermodynamics & KTG

Question

Hot water cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 1 minutes. In cooling from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ it will take (room temperature $=30^{\circ} \mathrm{C}$ )

Accepted Answer

Newton's law of cooling tells us how things cool down. It says:
$$ \frac{	heta_1-	heta_2}{\mathrm{t}} = \mathrm{K} \left( \frac{	heta_1+	heta_2}{2} - 	heta_0 ight) $$
Here, $	heta_1$ is the starting temperature, $	heta_2$ is the ending temperature, $t$ is the time taken, $K$ is a constant, and $	heta_0$ is the room temperature.
Step 1: Use the formula for cooling from $80^\circ$C to $60^\circ$C
Plug in the values:

$$ \frac{80-60}{60} = K \left( \frac{80+60}{2} - 30 ight) $$
This becomes:

$$ \frac{20}{60} = K (40) $$
Simplify:

$$ \frac{1}{3} = K (40) \qquad ...(i) $$
Step 2: Use the formula for cooling from $60^\circ$C to $50^\circ$C
Plug in the values:

$$ \frac{60-50}{t} = K \left( \frac{60+50}{2} - 30 ight) $$
This becomes:

$$ \frac{10}{t} = K (25) \qquad ...(ii) $$
Step 3: Divide equation (i) by equation (ii) to find $t$
Set up the equations:
$\frac{\frac{1}{3}}{K(40)} \div \frac{\frac{10}{t}}{K(25)}$
This simplifies to:
$\frac{1/3}{10/t} = \frac{40}{25}$
$\frac{1}{3} 	imes \frac{t}{10} = \frac{40}{25}$
Now, solve for $t$:
$\frac{t}{30} = \frac{40}{25}$
$t = \frac{40}{25} 	imes 30 = \frac{1200}{25} = 48$ seconds.

Answer

48 s

Answer

42 s

Answer

50 s

Answer

45 s

Thermodynamics & KTG

Hot water cools from 80^{\circ} \mathrm{C} to 60^{\circ} \mathrm{C} in 1 minutes. In cooling from 60^{\circ} \mathrm{C} to 50^{\circ} \mathrm{C} it will take (room temperature =30^{\circ} \mathrm{C} )

Choose the correct answer: