A body cools from 60.0 ∘ C to 59.8 ∘ C in 4 s. Find the time in seconds to cool from 45.0 ∘ C to 44.8 ∘ C in surroundings at 30.0 ∘ C (two decimals; Newton's law).

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferHard2 min

PhysicsHardnumerical

A body cools from $60.0^{\circ}$ C to $59.8^{\circ}$ C in 4 s. Find the time in seconds to cool from $45.0^{\circ}$ C to $44.8^{\circ}$ C in surroundings at $30.0^{\circ}$ C (two decimals; Newton's law).

Answer:

8.03

Solution:

The time is inversely proportional to the mean excess. First: $29.9$ , second: $14.9$ . $t = 4 \times \frac{29.9}{14.9} = 8.03$ s.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

⏱ 2m

Doubts & Discussion

Loading discussions...