A body cools from 80 °C to 64 °C in 5 minutes and same body cools from 80 °C to 52 °C in 10 minutes. What is the temperature of surrounding?
Temperature is a measure of the average kinetic energy of particles in a substance. It determines the direction of heat flow between objects, moving from higher to lower temperature. Measured in units like Celsius, Kelvin or Fahrenheit, temperature plays a crucial role in phase changes and physical processes.
Class 11 Physics Chapter 10 Thermal Properties of Matter focuses on heat transfer specific heat capacity thermal expansion and calorimetry. It covers conduction convection and radiation along with the concept of thermal equilibrium. The chapter also explains the laws of heat transfer and their applications which are important for the CBSE Exam 2024-25.
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We can solve this using Newton’s Law of Cooling, which states,
(dT/dt) = -k(T – Tₛ)
(dT/dt) is the rate of change of temperature
k is a constant
T is the temperature of the body
Tₛ is the temperature of the surroundings.
We are given that the body cools from 80°C to 64°C in 5 minutes and again from 80°C to 52°C in 10 minutes. Applying Newton’s Law, we can derive the temperature of the surroundings.
We can use this formula to find the temperatures:
ln[(T₁ – Tₛ) / (T₂ – Tₛ)] = k(t₂ – t₁)
For the first cooling process, where it cools from 80 °C to 64 °C in 5 minutes:
ln[(80 – Tₛ) / (64 – Tₛ)] = k × 5
For the second cooling (from 80 °C to 52 °C in 10 minutes):
ln[(80 – Tₛ) / (52 – Tₛ)] = k × 10
This system of equations solves for the value of Tₛ.
After solving, we find that the temperature of the surroundings is 25 °C.
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