How To Calculate Cooling Rate Of Water

Newton's Law of Cooling:

\[ \frac{dT}{dt} = - \frac{h A}{m c} (T - T_a) \]

Heat Transfer Coefficient (h):

W/m²K

Surface Area (A):

m²

Mass (m):

Specific Heat (c):

J/kg·K

Water Temperature (T):

Ambient Temperature (T_a):

Cooling Rate (dT/dt):

Unit Converter ▲

Unit Converter ▼

From:	To:

1. What is Newton's Law of Cooling?

Newton's Law of Cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. For water cooling, this provides a mathematical model to predict how quickly water temperature decreases over time.

2. How Does the Calculator Work?

The calculator uses Newton's Law of Cooling equation:

\[ \frac{dT}{dt} = - \frac{h A}{m c} (T - T_a) \]

Where:

\( \frac{dT}{dt} \) — Cooling rate (K/s)
\( h \) — Heat transfer coefficient (W/m²K)
\( A \) — Surface area (m²)
\( m \) — Mass of water (kg)
\( c \) — Specific heat capacity of water (J/kg·K)
\( T \) — Water temperature (K)
\( T_a \) — Ambient temperature (K)

Explanation: The negative sign indicates temperature decrease over time. The cooling rate depends on the temperature difference and the system's thermal properties.

3. Importance of Cooling Rate Calculation

Details: Calculating cooling rates is essential for thermal management systems, industrial processes, food safety, and understanding heat transfer in various applications from engineering to environmental science.

4. Using the Calculator

Tips: Enter all parameters in the specified units. Ensure temperatures are in Kelvin. All values must be positive, with ambient temperature typically lower than water temperature for cooling to occur.

5. Frequently Asked Questions (FAQ)

Q1: What is the typical heat transfer coefficient for water?
A: For natural convection in air, h ranges from 5-25 W/m²K. For forced convection, it can be much higher (50-1000 W/m²K).

Q2: Why use Kelvin instead of Celsius?
A: Kelvin is the SI unit for temperature and ensures proper calculation of temperature differences in thermodynamic equations.

Q3: What is the specific heat of water?
A: Approximately 4186 J/kg·K at room temperature, though it varies slightly with temperature.

Q4: Does this equation work for heating as well?
A: Yes, the same equation applies to heating when T_a > T, resulting in a positive dT/dt (warming rate).

Q5: What are the limitations of Newton's Law of Cooling?
A: It assumes constant heat transfer coefficient and works best for small temperature differences. For large temperature ranges, more complex models may be needed.