Interactive Thermal Physics Laboratory
Newton's Law of Cooling
Newton's Law of Cooling states that the rate of loss of heat from a body is directly proportional to the difference in temperature between the body and its surroundings.
Newton's Law of Cooling Simulator
1. What is Newton's Law of Cooling?
Newton's Law of Cooling describes the rate at which an exposed object changes temperature through heat transfer with its surroundings. According to the law, the rate of temperature drop is directly proportional to the difference between the object's temperature and the temperature of the surrounding medium (ambient temperature).
This law applies primarily to heat transfer driven by convection (like a breeze cooling a hot bowl of soup) and conduction, under the assumption that the temperature difference is relatively small.
2. Mathematical Formulas
Newton's Law of Cooling can be formulated in two ways:
Differential Form
The instantaneous rate of change of temperature with respect to time is:
Where:
- \(\frac{dT}{dt}\) = Rate of cooling (degrees per unit time, e.g. °C/minute)
- \(k\) = Cooling constant (a positive constant, min⁻¹ or s⁻¹)
- \(T\) = Instantaneous temperature of the body (°C or Kelvin)
- \(T_s\) = Temperature of the surroundings/ambient medium (°C or Kelvin)
Integrated Exponential Form
Integrating the differential equation over time yields the formula for temperature at any time \(t\):
Where:
- \(T(t)\) = Temperature of the body at time \(t\)
- \(T_0\) = Initial temperature of the body at time \(t = 0\)
- \(e\) = Euler's number \(\approx 2.71828\)
The integrated form shows that the temperature difference (\(T - T_s\)) decreases exponentially over time.
3. Solved Examples
- Recall the exponential solution to Newton's Law of Cooling: T(t) = Tₛ + (T₀ - Tₛ) * e^(-kt).
- Identify parameters: T(10) = 65.0°C, T₀ = 85.0°C, Tₛ = 25.0°C, t = 10.0 minutes.
- Substitute values: 65.0 = 25.0 + (85.0 - 25.0) * e^(-k * 10).
- Simplify: 40.0 = 60.0 * e^(-10k) => e^(-10k) = 40.0 / 60.0 = 2/3 ≈ 0.6667.
- Take the natural logarithm of both sides: -10k = ln(0.6667) ≈ -0.4055.
- Solve for k: k = -0.4055 / -10.0 = 0.0405 min⁻¹.
- Verify: The cooling constant is approximately 0.0405 per minute. This means the temperature difference drops by about 4.05% every minute.
- Identify parameters: T₀ = 80.0°C, Tₛ = 20.0°C, k = 0.030 min⁻¹, t = 15.0 min.
- Write the formula: T(t) = Tₛ + (T₀ - Tₛ) * e^(-kt).
- Substitute values: T(15) = 20.0 + (80.0 - 20.0) * e^(-0.030 * 15).
- Calculate the exponent: -0.030 * 15 = -0.45.
- Evaluate exponential term: e^(-0.45) ≈ 0.6376.
- Calculate final temperature: T(15) = 20.0 + 60.0 * 0.6376 = 20.0 + 38.26 = 58.26°C.
- Verify: After 15 minutes, the soup has cooled down to approximately 58.3°C, which is a warm and comfortable eating temperature.
- Recall Newton's Law of Cooling in differential form: dT/dt = -k * (T - Tₛ).
- Identify parameters: T = 70.0°C, Tₛ = 22.0°C, k = 0.050 min⁻¹.
- Calculate the temperature difference: T - Tₛ = 70.0°C - 22.0°C = 48.0°C.
- Substitute values: dT/dt = -0.050 min⁻¹ * 48.0°C.
- Calculate: dT/dt = -2.40°C/minute.
- Verify: The negative sign indicates cooling. At the exact moment the temperature is 70°C, the water is cooling at a rate of 2.4°C per minute.
4. Practice Questions
5. Frequently Asked Questions (FAQs)
What is Newton's Law of Cooling?
It is a physical law stating that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature of its surroundings.
What is the formula for Newton's Law of Cooling?
The differential form is dT/dt = -k(T - Tₛ). The integrated exponential form is T(t) = Tₛ + (T₀ - Tₛ)e^(-kt), where T(t) is temperature at time t, Tₛ is room temp, T₀ is starting temp, and k is the cooling constant.
Why is the cooling curve exponential?
Because the rate of cooling decreases as the temperature of the object drops. Since the rate of change is proportional to the value itself, it integrates to a natural exponential decay function.
What are the units of the cooling constant (k)?
The units are reciprocal time, such as per second (s⁻¹), per minute (min⁻¹), or per hour (hr⁻¹).
How does airflow affect the cooling rate?
Moving air increases forced convection, removing warm air from around the hot object and replacing it with cooler air. This increases the heat transfer coefficient, leading to a larger cooling constant (k) and faster cooling.
Does a larger volume of water cool faster or slower?
A larger volume (and mass) of water cools slower. While it has slightly more surface area, its thermal mass (m·c) increases much more than its surface area, which reduces the cooling constant (k = h·A / (m·c)) and slows the temperature drop.
What is the difference between heat loss and temperature drop?
Heat loss (dQ/dt) is the energy leaving the object in Watts (J/s). Temperature drop (dT/dt) is the actual drop in temperature in °C/s. They are related by dQ/dt = m·c·(dT/dt), where m is mass and c is specific heat capacity.
Does the law apply to heating as well?
Yes. It can be written as dT/dt = k(Tₛ - T) for heating. A cold object (like a soda can taken out of the fridge) warms up rapidly at first, then more slowly as its temperature approaches room temperature.
What is convective heat transfer?
It is the transfer of heat between a solid surface and a moving fluid (like air or water). Convection is the primary driver of cooling in hot mugs, plates, and open liquids.
Who formulated this law?
Sir Isaac Newton formulated and published it in 1701 in a short paper on temperature scales and heat conduction.